A Brief Review of Real and Functional Analysis

Abstract

This chapter is meant to provide a very quick review of the real and functional analysis results that will be most frequently used afterwards. Missing proofs can be found in most classical works dealing with these questions.

Appendix: The Topologies of \(\mathcal {D}\) and \(\mathcal {D}'\)

In the literature of applied partial differential equations, it is customary not to dwell too much on the description of the topologies of $\mathcal {D}(\Omega )$ and $\mathcal {D}'(\Omega )$. It is true that there is no need to know them in detail in order to work efficiently. The sequential properties sketched above are amply sufficient. There is thus no real drawback in not reading the rest of the present section.

On the other hand, it is quite legitimate to be curious with regard to these topologies, without being willing to read through the theory of abstract topological vector spaces, which is rather imposing, see [8, 63, 68]. In effect, we are dealing with topological vector spaces that are not normed spaces, but are naturally equipped with much more sophisticated topologies.

A topological vector space on ${\mathbb R}$, or more generally on a topological field $\mathbb {K}$, is a $\mathbb {K}$-vector space E equipped with a topology which is such that the addition is continuous from E × E to E and the scalar multiplication is continuous from $\mathbb {K}\times E$ to E, both product spaces being equipped with their product topology. We will stick with $\mathbb {K}={\mathbb R}$ from now on.

Let us start with the concept of Fréchet space. A seminorm on a vector space on ${\mathbb R}$ is a mapping with nonnegative values that is absolutely homogeneous and satisfies the triangle inequality, i.e., is subadditive.

Let E be an ${\mathbb R}$-vector space and $(p_n)_{n\in {\mathbb N}}$ a countable, nondecreasing family of seminorms on E such that for all u ≠ 0 in E, there exists $n\in {\mathbb N}$ with p _n (u) > 0. For all $n\in {\mathbb N}$ and α > 0, we let

$$\displaystyle \begin{aligned} V_{n,\alpha}(u)=\{v\in E; p_n(v-u)<\alpha\}. \end{aligned}$$

We define a family $\mathcal {O}$ of subsets of E by saying that

$$\displaystyle \begin{aligned} U\in \mathcal{O}\text{ if and only if } \forall u\in U, \exists n\in{\mathbb N},\exists\alpha\in {\mathbb R}_+^*, V_{n,\alpha}(u)\subset U. \end{aligned} $$

This is reminiscent of the way a topology arises from a distance in a metric space, and the sets V _n,α(u) are meant to evoke neighborhoods of u, except that we do not have one distance, but a whole family of seminorms.

Proposition 1.13

The family $\mathcal {O}$ is a topology on E which makes it a topological vector space. This topology is said to be generated by the family of seminorms . This topology is metrizable and the mapping $E\times E\to {\mathbb R}_+$ ,

$$\displaystyle \begin{aligned} d(u,v)=\sum_{n=0}^\infty 2^{-n}\min(1,p_n(u-v)), \end{aligned} $$

(1.5)

is a distance that also generates the topology.

Proof

Let us check that the axioms of a topology are satisfied. Trivially, $E\in \mathcal {O}$ and $\emptyset \in \mathcal {O}$, since in the latter case, the condition to be met is empty.

Let (U _i )_i=1,…,k be a finite family of elements of $\mathcal {O}$ and $U=\cap _{i=1}^k U_i$.⁹ Let us take u ∈ U. By definition, for all i = 1, …, k, there exists $n_i\in {\mathbb N},\alpha _i>0$ such that $V_{n_i,\alpha _i}(u)\subset U_i$. We set $n=\max \{n_i\}\in {\mathbb N}$ and $\alpha =\min \{\alpha _i\}>0$. Now the sequence p _n is nondecreasing, so that the inequalities

$$\displaystyle \begin{aligned}p_{n_i}(v-u)\le p_n(v-u)<\alpha\le\alpha_i, \end{aligned} $$

show that $V_{n,\alpha }(u)\subset V_{n_i,\alpha _i}(u)\subset U_i$ for all i. Consequently, V _n,α(u) ⊂ U, which implies that $U\in \mathcal {O}$.

Let now (U _λ )_{λ ∈ Λ} be an arbitrary family of elements of $\mathcal {O}$ and U = ∪_{λ ∈ Λ} U _λ . Let us take u ∈ U. By definition, there exists λ ∈ Λ such that u ∈ U _λ . We can thus choose an n and an α such that V _n,α(u) ⊂ U _λ . Trivially, U _λ  ⊂ U and thus $U\in \mathcal {O}$.

We are thus assured that we are dealing with a topology. Let us check that this topology is a topological vector space topology, i.e., that the vector space operations are continuous.

For the addition, let us be given v ₁ and v ₂ in E, and let u = v ₁ + v ₂. We consider an open set U that contains u. There exist n and α such that V _n,α(u) ⊂ U. We claim that V _n,α∕2(v ₁) + V _n,α∕2(v ₂) ⊂ V _n,α(u), which implies the continuity of the addition at point (v ₁, v ₂). Indeed, if w ₁ ∈ V _n,α∕2(v ₁) and w ₂ ∈ V _n,α∕2(v ₂), then

$$\displaystyle \begin{aligned}\displaystyle p_n(w_1+w_2-u)=p_n(w_1-v_1+w_2-v_2)\\\displaystyle \le p_n(w_1-v_1)+p_n(w_2-v_2)<\frac\alpha2+\frac\alpha2=\alpha \end{aligned} $$

by the triangle inequality.

For the scalar multiplication, we note first that for all λ, μ, u, and v,

$$\displaystyle \begin{aligned} \mu v-\lambda u=\mu(v-u)+(\mu-\lambda)u. \end{aligned}$$

Let U be an open set containing λu and V _n,α(λu) be an open neighborhood of λu that is included in U. We wish to find η > 0 and β > 0 such that if (μ, v) ∈ ]λ − η, λ + η[ × V _n,β(u), then μv ∈ V _n,α(λu).

It follows from the above decomposition that

$$\displaystyle \begin{aligned} p_n(\mu v-\lambda u)\le|\mu|p_n(v-u)+|\mu-\lambda|p_n(u) \end{aligned}$$

by absolute homogeneity of the seminorm and the triangle inequality. There are two cases. First of all, if p _n (u) > 0, an obvious choice for η is then $\eta =\frac {\alpha }{2 p_n(u)}$, which yields

$$\displaystyle \begin{aligned} |\mu-\lambda|p_n(u)<\frac\alpha2. \end{aligned}$$

If p _n (u) = 0, then we take η = 1, and the above estimate still holds true. Then, since |μ| < |λ| + η, we can take $\beta =\frac \alpha {2(|\lambda |+\eta )}$ and obtain

$$\displaystyle \begin{aligned} |\mu|p_n(v-u)<\frac\alpha2, \end{aligned}$$

thus establishing the continuity of scalar multiplication.

Let us finally show that this topological vector space topology is metrizable. It is not difficult to check that d defined by formula (1.5) is a distance on E. This is where the hypothesis that for all u ≠ 0, there exists n such that p _n (u) > 0, comes into play, in order to ensure that d(v, w) = 0 if and only if v = w. To conclude, we must show that for all u ∈ E, every open set of $\mathcal {O}$ containing u contains an open ball of d also containing u, and conversely. Since all the quantities involved are translation invariant, it is enough to prove this for u = 0. We denote by B(0, α) the open ball of d centered at 0 and of radius α.

Let us thus be given an open set U of $\mathcal {O}$ that contains 0. By definition, there exists $n\in {\mathbb N}$ and α > 0 such that V _n,α(0) ⊂ U. We want to find β > 0 such that B(0, β) ⊂ V _n,α(0). We notice that it is enough to consider the case α < 1, since $V_{n,\alpha }(0)\subset V_{n,\alpha '}(0)$ as soon as α ≤ α′. We take β = 2⁻ⁿ α. If v ∈ B(0, β), then

$$\displaystyle \begin{aligned} 2^{-n}\min (1,p_n(v))\le d(0,v)<2^{-n}\alpha. \end{aligned}$$

Since α < 1, the minimum in the left-hand side is not 1 and it follows that p _n (v) < α, or in other words v ∈ V _n,α(0), hence B(0, β) ⊂ V _n,α(0).

Conversely, let us be given β > 0 and consider the ball B(0, β). For all $n\in {\mathbb N}$, there holds

$$\displaystyle \begin{aligned} d(0,v)=\sum_{k=0}^n2^{-k}\min(1,p_k(v))+\sum_{k=n+1}^\infty2^{-k}\min(1,p_k(v)). \end{aligned}$$

Now $\min (1,p_k(v))\le 1$, therefore

$$\displaystyle \begin{aligned} \sum_{k=n+1}^\infty2^{-k}\min(1,p_k(v))\le 2^{-n}. \end{aligned}$$

We pick n large enough so that 2⁻ⁿ < β∕2. Since $\min (1,p_k(v))\le p_k(v)$ and the sequence of seminorms is nondecreasing, we have

$$\displaystyle \begin{aligned} \sum_{k=0}^n2^{-k}\min(1,p_k(v))\le 2p_n(v). \end{aligned}$$

We then choose α = β∕4. It follows that if v ∈ V _n,α(0), then

$$\displaystyle \begin{aligned} d(0,v)\le 2p_n(v)+\beta/2<\beta, \end{aligned}$$

that is to say V _n,α(0) ⊂ B(0, β). □

More generally, the topological vector spaces whose topology is generated by an arbitrary family of seminorms, i.e., not necessarily a countable family, are called locally convex topological vector spaces. It is (almost¹⁰) clear that the countability of the seminorm family plays no role in the fact that we are dealing with a topological vector space. The countability only comes into play for metrizability. An equivalent characterization is the existence of a basis of convex neighborhoods of 0, hence the name. The equivalence uses the gauge or Minkowski functional of a convex set , see Chap. 2. All the topological vector spaces considered here fall into this category.

Definition 1.3

We say that a vector space E equipped with a countable family of seminorms as above is a Fréchet space if it is complete for the distance (1.5).

Remark 1.3

Fréchet spaces provide an example of the usefulness of the concept of metrizable space and the subtle distinction made with metric spaces: their topology is naturally defined using neighborhoods. It so happens that there is a distance that generates the same topology, but this distance is not especially natural. Besides, there are many other distances that are equivalent to it. When we work in a Fréchet space, we rarely use the distance explicitly. To the contrary, seminorms and associated neighborhoods will be used. Note that a Fréchet space being metrizable and complete, it is a Baire space. □

Definition 1.4

A subset A of a topological vector space E is said to be bounded if for any neighborhood V of 0, there exists a scalar λ such that A ⊂ λV (we say that A is absorbed by any neighborhood of 0).

Caution

The concept of bounded subset just introduced is not a metric concept, but a topological vector space concept. As a matter of fact, the distance d defined above in the case of a countable family of seminorms is itself bounded. Clearly, the diameter of E is less than 2, which is not what we have in mind for a bounded subset of a topological vector space. The two concepts however coincide in the case of a normed vector space when we consider the distance that is canonically associated with the norm, d(u, v) = ∥u − v∥.

Bounded subsets of E are easily characterized in terms of the seminorms used to define the topology.

Proposition 1.14

Let E be a topological vector space the topology of which is generated by a family of seminorms $(p_n)_{n\in {\mathbb N}}$ . A subset A of E is bounded if and only if for all n, there exists a constant λ _n such that for all u ∈ A, p _n (u) < λ _n .

Proof

Let A be a bounded subset of E. For all n, V _n,1(0) is a neighborhood of 0. Thus, by definition, for all n, there exists λ _n such that $A\subset \lambda _n V_{n,1}(0)=V_{n,\lambda _n}(0)$, which means that for all u ∈ A, p _n (u) < λ _n .

Conversely, let A satisfy the condition of the Proposition. Let U be a neighborhood of 0. It contains a neighborhood of the form V _n,α(0), for some n and α. By hypothesis, for all u ∈ A, p _n (u) < λ _n . In particular λ _n  > 0. Consequently, p(αu∕λ _n ) < α, by absolute homogeneity. This says that $\frac {\alpha }{\lambda _n}u\in V_{n,\alpha }(0)$, which in turn readily implies that $A\subset \frac {\lambda _n}\alpha V_{n,\alpha }(0)\subset \frac {\lambda _n}\alpha U$. □

A subset of E is thus bounded if and only if it is bounded for every seminorm. Here too, the countability of the family plays no role in the above characterization of bounded subsets.

If the sequence of seminorms is stationary, i.e., if there exists $n_0\in {\mathbb N}$ such that $p_n=p_{n_0}$ for all n ≥ n ₀, then it is easily checked that $p_{n_0}$ is a norm that generates the topology. We have thus only introduced a possibly new object compared to normed spaces, or Banach spaces in the complete case, if the sequence p _n is not stationary. The latter condition is however not sufficient to ensure that a given Fréchet space is not normable.

Proposition 1.15

Let E be a topological vector space, the topology of which is generated by a nondecreasing family of seminorms $(p_n)_{n\in {\mathbb N}}$ such that for all n, there exists m > n such that p _m is not equivalent to p _n . Then the topology of E is not normable.

Proof

We have p _n  ≤ p _m , therefore saying that p _m is not equivalent to p _n means that $\sup \{p_m(u);u\in E,p_n(u)<1\}=+\infty$. If a topological vector space is normable, it has a bounded set with nonempty interior, namely the unit ball of a norm that generates the topology. We are thus going to show that every bounded set of E has empty interior.

Let thus A be a bounded subset of E and λ _n , $n\in {\mathbb N}$, the scalars that express its boundedness in terms of the seminorms. Let U be an open set included in A and assume that U is nonempty. It thus contains a neighborhood V _n,α(u) for some triple (n, α, u) with α > 0. We take m > n as above. For all v ∈ V _n,α(u), we see that $w=\frac {v-u}\alpha \in V_{n,1}(0)$, and conversely if w ∈ V _n,1(0), then v = u + αw ∈ V _n,α(u). By hypothesis, there exists w ∈ V _n,1(0) such that $p_m(w)\ge \frac {\lambda _m+p_m(u)}\alpha$. It follows that p _m (v) ≥ αp _m (w) − p _m (u) ≥ λ _m by the triangle inequality, with v ∈ V _n,α(u) ⊂ U ⊂ A. This is a contradiction, therefore U is empty. □

Conversely, it is not hard to see that if all seminorms are equivalent starting from a certain rank, then the topology is normable by one of these equivalent seminorms.

Since the topology of a Fréchet space is metrizable, it can also be worked out in terms of convergent sequences. These sequences are very easily described.

Proposition 1.16

Let E be a topological vector space with a countable family of seminorms as above. A sequence u _k tends to u in the sense of E if and only if, p _n (u _k  − u) → 0 when k → +∞, for all $n\in {\mathbb N}$.

Proof

A sequence u _k tends to u if and only if, for any neighborhood V of u, there exists k ₀ such that u _k  ∈ V for all k ≥ k ₀. This happens if and only if for all n and α > 0, there exists k ₀ such that u _k  ∈ V _n,α(u), that is to say p _n (u _k  − u) < α, for all k ≥ k ₀. □

Of course, for such a sequence d(u _k , u) → 0 and conversely, which is a simple exercise when we do not know yet that both topologies coincide.

Let us stop here with generalities about Fréchet spaces and introduce our main example in the context of distributions.

Proposition 1.17

Let Ω be an open subset of ${\mathbb R}^d$ and K a compact subset of Ω with nonempty interior. The space

$$\displaystyle \begin{aligned} \mathcal{D}_K(\Omega)=\{\varphi\in C^\infty(\Omega);\mathop{\mathrm{supp}} u\subset K\} \end{aligned}$$

of indefinitely differentiable functions with support in K, equipped with the family of seminorms

$$\displaystyle \begin{aligned} p_n(\varphi)=\max_{|\gamma|\le n, x\in K}|\partial^\gamma\varphi(x)|, \end{aligned} $$

(1.6)

is a Fréchet space.

Proof

The family of seminorms is clearly nondecreasing. Moreover, p ₀ is a norm, so that p ₀(φ) > 0 as soon as φ ≠ 0. The only difficulty is the completeness.

Let thus $\varphi _k\in \mathcal {D}_K(\Omega )$ be a Cauchy sequence . We notice that p _n actually is the norm on $C^n_K(\Omega )$. Therefore, if φ _k is Cauchy in $\mathcal {D}_K(\Omega )$, it is a fortiori Cauchy in $C^n_K(\Omega )$ for all n. Now $C^n_K(\Omega )$ is complete, and $C^{n+1}_K(\Omega )\hookrightarrow C^n_K(\Omega )$. It follows that φ _k converge in $C^n_K(\Omega )$ toward some φ, which is the same for all n. Hence, $\varphi \in \mathcal {D}_K(\Omega )$, and according to Proposition 1.16, the sequence φ _k converges to φ in $\mathcal {D}_K(\Omega )$. □

Let us remark in passing that Proposition 1.16 translates in this particular case into the fact that a sequence converges in $\mathcal {D}_K(\Omega )$ if and only if all its partial derivatives at all orders converge uniformly on K.

The space $\mathcal {D}_K(\Omega )$ is not normable because its family of seminorms satisfies the hypothesis of Proposition 1.15. There is no norm that can generate its Fréchet space topology, at least we have not worked for nothing. This can also be seen as a consequence of the next proposition, which can be slightly surprising at first when we only know about infinite dimensional normed spaces.

Proposition 1.18

Bounded closed subsets of $\mathcal {D}_K(\Omega )$ are compact.

Proof

Let A be a bounded subset of $\mathcal {D}_K(\Omega )$. For all n, there thus exists λ _n such that

$$\displaystyle \begin{aligned} \forall \varphi\in A,\quad p_n(\varphi)\le\lambda_n. \end{aligned}$$

By the mean value inequality , this implies that ∂ ^γ A is an equicontinuous family in $C^0_K(\Omega )$ for all multi-indices |γ|≤ n − 1, and that max _K |∂ ^γ φ|≤ λ _n for all φ ∈ A. Since K is compact, we can apply Ascoli’s theorem to deduce that all these sets are relatively compact in $C^0_K(\Omega )$.

Let us now take a sequence in A. Using the above remark, we extract a subsequence that converges in all $C^n_K(\Omega )$, $n\in {\mathbb N}$, by the diagonal argument . The set A is thus relatively compact. □

Of course, in a separated topological vector space, all compact subsets are bounded and closed. A space that has the property of Proposition 1.18 and is reflexive is called a Montel space. Because of the Riesz theorem , an infinite dimensional normed space is not a Montel space. Now, since K has nonempty interior, $\mathcal {D}_K(\Omega )$ is patently infinite dimensional, hence is not normable, as we already noticed before.¹¹

Be careful that this does not mean that $\mathcal {D}_K(\Omega )$ is locally compact. In fact, a more general version of Riesz’s theorem states that a separated topological vector space is locally compact if and only if it is finite dimensional, see [63]. Simply, here the closed bounded sets, which are the compact sets of $\mathcal {D}_K(\Omega )$, all have empty interior and no nonempty open set is relatively compact.

It should finally be kept in mind that since $\mathcal {D}_K(\Omega )$ is infinite dimensional, it can be equipped with several different reasonable topologies. The topology we have described up to now is called the strong topology of $\mathcal {D}_K(\Omega )$.

Let us now talk about the dual space of $\mathcal {D}_K(\Omega )$, denoted $\mathcal {D}^{\prime }_K(\Omega )$.

Proposition 1.19

A linear form T on $\mathcal {D}_K(\Omega )$ is continuous if and only if there exists $n\in {\mathbb N}$ and $C\in {\mathbb R}$ such that

$$\displaystyle \begin{aligned} \forall \varphi\in\mathcal{D}_K(\Omega), \quad |\langle T,\varphi\rangle|\le Cp_n(\varphi), \end{aligned} $$

(1.7)

and if and only if, for any sequence φ _k  → φ in $\mathcal {D}_K(\Omega )$ , 〈T, φ _k 〉→〈T, φ〉.

Proof

The second characterization is trivial since $\mathcal {D}_K(\Omega )$ is metrizable. For the first characterization, we start with noticing that it is enough to prove the continuity of T at 0 by linearity. Let T be a linear form that satisfies (1.7). Since φ _k  → 0 implies that p _n (φ _k ) → 0 for all n, we clearly have 〈T, φ _k 〉→ 0, hence T is continuous.

Conversely, let us take $T\in \mathcal {D}^{\prime }_K(\Omega )$. This is a continuous mapping from $\mathcal {D}_K(\Omega )$ into ${\mathbb R}$, therefore the preimage of any open set of ${\mathbb R}$ is an open set of $\mathcal {D}_K(\Omega )$. In particular, since 0 ∈ T ⁻¹(]−1, 1[), there exists $n\in {\mathbb N}$ and α > 0 such that V _n,α(0) ⊂ T ⁻¹(]−1, 1[). In other words, this means that if p _n (φ) < α, then |〈T, φ〉| < 1. Now, if φ ≠ 0, then $p_n\bigl (\frac {\alpha \varphi }{2p_n(\varphi )}\bigr )<\alpha$ by absolute homogeneity. It follows that for all nonzero φ, $\big |\bigl \langle T,\frac {\alpha \varphi }{2p_n(\varphi )}\bigr \rangle \big |<1$. This implies that $|\langle T,\varphi \rangle |\le \frac 2\alpha p_n(\varphi )$ for all φ, including φ = 0 for which the latter inequality obviously holds. □

Note the little trick of switching from strict inequalities to a non-strict one in the end in order to accommodate the case φ = 0. With a little more work, $\frac 2\alpha$ can be replaced by $\frac 1\alpha$.

Which topology are we going to use on $\mathcal {D}^{\prime }_K(\Omega )$? Once again, there are several choices. We will only be interested here in the weak-star topology . Just like in the case of the dual space of a normed vector space, cf. Sect. 1.7, this is the coarsest topology, that is to say the one with as little open sets as possible, which makes all the mappings of the form T↦〈T, φ〉 continuous, with φ arbitrary in $\mathcal {D}_K(\Omega )$.

Let us focus for a moment on this concept of “the coarsest topology having this or that property,” from an abstract viewpoint.

Proposition 1.20

Let X be a set and $\mathcal {A}\subset \mathcal {P}(X)$ be a set of subsets of X. There exists a unique topology on X which is the coarsest of all topologies containing $\mathcal {A}$ . This topology is called the topology generated by $\mathcal {A}$. It consists in all arbitrary unions of finite intersections of elements of $\mathcal {A}$.

Proof

A topology on X is an element of $\mathcal {P}(\mathcal {P}(X))$ with special properties. It thus makes sense to consider families of topologies on X. It is then fairly obvious that the intersection of such a nonempty family of topologies is a topology on X. Now $\mathcal {A}$ is also an element of $\mathcal {P}(\mathcal {P}(X))$, and the discrete topology $\mathcal {P}(X)$ contains $\mathcal {A}$. Consequently, the family of topologies containing $\mathcal {A}$ is nonempty, and its intersection is the smallest possible topology containing $\mathcal {A}$.

Let us describe this topology more explicitly. If it contains $\mathcal {A}$, it must contain all finite intersections of elements of $\mathcal {A}$, due to closure under finite intersections. Due to closure under arbitrary unions, it must then contain arbitrary unions of such finite intersections. It is thus enough to show that the set of arbitrary unions of finite intersections of elements of $\mathcal {A}$ is a topology.

Let $\mathcal {O}$ be this set. It obviously contains ∅ and X, and is closed under arbitrary unions. The only (small) difficulty is closure under finite intersections. It is enough to consider the case of two elements U ₁ and U ₂ of $\mathcal {O}$, the general case follows by induction on the number of elements in the family.

Let us thus be given U ₁ and U ₂ such that there are two sets of indices Λ ₁ and Λ ₂, and for each λ ∈ Λ _i , an integer p _λ such that we can write¹²

$$\displaystyle \begin{aligned}U_1=\bigcup_{\lambda\in\varLambda_1}\Bigl(\bigcap_{k=1}^{p_{\lambda}}A_{\lambda,k}\Bigr),\quad U_2=\bigcup_{\mu\in\varLambda_2}\Bigl(\bigcap_{l=1}^{p_{\mu}}A_{\mu,l}\Bigr), \end{aligned}$$

where A _λ,k and A _μ,l belong to $\mathcal {A}$. We want to show that $U_1\cap U_2\in \mathcal {O}$. Let us set Λ = Λ ₁ × Λ ₂ and

$$\displaystyle \begin{aligned}V=\bigcup_{(\lambda,\mu)\in\varLambda}\Bigl(\Bigl(\bigcap_{k=1}^{p_{\lambda}}A_{\lambda,k}\Bigr)\bigcap\Bigl(\bigcap_{l=1}^{p_{\mu}}A_{\mu,l}\Bigr)\Bigr), \end{aligned}$$

so that $V\in \mathcal {O}$. Let x ∈ U ₁ ∩ U ₂. There exist λ ∈ Λ ₁ and μ ∈ Λ ₂ such that $x\in \bigcap _{k=1}^{p_{\lambda }}A_{\lambda ,k}$ and $x\in \bigcap _{l=1}^{p_{\mu }}A_{\mu ,l}$. In other words, $x\in \bigl (\bigcap _{k=1}^{p_{\lambda }}A_{\lambda ,k}\bigr )\bigcap \bigl (\bigcap _{l=1}^{p_{\mu }}A_{\mu ,l}\bigr )$. This shows that U ₁ ∩ U ₂ ⊂ V .

Conversely, let us take x ∈ V . Then, there exists (λ, μ) ∈ Λ such that $x\in \bigl (\bigcap _{k=1}^{p_{\lambda }}A_{\lambda ,k}\bigr )\bigcap \bigl (\bigcap _{l=1}^{p_{\mu }}A_{\mu ,l}\bigr )$, that is to say $x\in \bigcap _{k=1}^{p_{\lambda }}A_{\lambda ,k}$ and $x\in \bigcap _{l=1}^{p_{\mu }}A_{\mu ,l}$, that is to say x ∈ U ₁ and x ∈ U ₂. This shows that V ⊂ U ₁ ∩ U ₂. With the previous inclusion, it follows that V = U ₁ ∩ U ₂, hence the intersection of two elements of $\mathcal {O}$ belongs to $\mathcal {O}$. We conclude by induction on the number of elements of $\mathcal {O}$ to be intersected with each other. □

Definition 1.5

Let X be a set, (X _λ )_{λ ∈ Λ} a family of topological spaces and for each λ, a mapping f _λ : X → X _λ . The coarsest topology on X that makes all the mappings f _λ continuous is called the projective topology or initial topology with respect to the family (X _λ , f _λ )_{λ ∈ Λ}.

This topology exists and is unique. Indeed, it is simply the topology generated by the family of sets $f_\lambda ^{-1}(U_\lambda )$ where λ range over Λ and for each such λ, U _λ range over the open sets of X _λ . A basis for the topology—that is to say a family of sets that generate the open sets by arbitrary unions—is given by sets of the form $\bigcap _{k=1}^{p}f_{\lambda _k}^{-1}(U_{\lambda _k})$ where $U_{\lambda _k}$ is an open set of $X_{\lambda _k}$, according to Proposition 1.20. It follows that a mapping f : Y → X from a topological space Y into X equipped with the projective topology is continuous if and only if for all λ ∈ Λ, f _λ  ∘ f is continuous from Y to X _λ . Finally, why look for the coarsest topology in this case? This is because it lives on X which is the domain of all f _λ . Adding sets to a topology on X can only improve the continuity status of these mappings, so the challenge is really to remove as many of them as possible.

The convergent sequences for this topology are also very simple.

Proposition 1.21

Let x _n be a sequence in X equipped with the projective topology with respect to the family (X _λ , f _λ )_{λ ∈ Λ} . Then x _n  → x in X if and only if f _λ (x _n ) → f _λ (x) in X _λ , for all λ ∈ Λ.

Proof

Let us assume that x _n  → x. Since each f _λ is continuous, it follows that f _λ (x _n ) → f _λ (x).

Conversely, let us assume that f _λ (x _n ) → f _λ (x) for all λ ∈ Λ. Let us take a neighborhood of x for the projective topology, of the form $\bigcap _{k=1}^{p}f_{\lambda _k}^{-1}(U_{\lambda _k})$. By hypothesis, for all 1 ≤ k ≤ p, there exists an integer n _k such that $f_{\lambda _k}(x_n)\in U_{\lambda _k}$ for all n ≥ n _k . Let us set $n_0=\max \{n_1,\ldots ,n_p\}$. We thus see that $x_n\in \bigcap _{k=1}^{p}f_{\lambda _k}^{-1}(U_{\lambda _k})$ for all n ≥ n ₀. This shows that x _n  → x for the projective topology. □

Let us apply all this to $\mathcal {D}^{\prime }_K(\Omega )$. The weak-star topology is nothing but the projective topology with respect to $\varLambda =\mathcal {D}_K(\Omega )$, λ = φ, $X_\varphi ={\mathbb R}$ and f _φ (T) = 〈T, φ〉. According to Proposition 1.20, a neighborhood basis for 0 is given by sets of the form $\bigcap _{k=1}^n\{T\in \mathcal {D}^{\prime }_K(\Omega );|\langle T,\varphi _k\rangle |<\varepsilon \}$. This neighborhood basis makes it very easy to check that we are dealing with a topological vector space topology, i.e., that the vector space operations are continuous. Furthermore, we see that this neighborhood basis is associated with the seminorms p(T) =max_k≤n|〈T, φ _k 〉|, hence it is a locally convex topology. A sequence T _n converges to T for the weak-star topology if and only if 〈T _n , φ〉→〈T, φ〉 for all $\varphi \in \mathcal {D}_K(\Omega )$.¹³

Let us now deal with the space $\mathcal {D}(\Omega )$, a little faster. First of all, $\mathcal {D}(\Omega )$ is actually a vector space. Indeed, $\mathop {\mathrm {supp}}(\varphi +\psi )\subset \mathop {\mathrm {supp}}(\varphi )\cup \mathop {\mathrm {supp}}(\psi )$ and $\mathop {\mathrm {supp}}(\lambda \varphi )\subset \mathop {\mathrm {supp}}(\varphi )$ which are compact subsets of Ω.

We have to start again with some abstraction.

Proposition 1.22

Let X be a set and $(\mathcal {O}_\lambda )_{\lambda \in \varLambda }$ a nonempty family of topologies on X. There exists a unique topology $\mathcal {O}$ on X which is the finest of all the topologies included in each $\mathcal {O}_\lambda$.

Proof

It is enough to take $\mathcal {O}=\bigcap _{\lambda \in \varLambda }\mathcal {O}_\lambda$ which is obviously a topology, hence by construction the largest in the sense of inclusion that is contained in each $\mathcal {O}_\lambda$. □

A set is thus an open set of $\mathcal {O}$ if and only if it is an open set of $\mathcal {O}_\lambda$ for all λ ∈ Λ.

Definition 1.6

Let X be a set, (X _λ )_{λ ∈ Λ} a family of topological spaces and for each λ, a mapping f _λ : X _λ  → X. The finest topology on X that makes all the f _λ continuous is called the inductive topology or final topology with respect to the family (X _λ , f _λ )_{λ ∈ Λ}.

This topology is well defined. Indeed, let

$$\displaystyle \begin{aligned}\mathcal{O}_\lambda=\{U\subset X; f_\lambda^{-1}(U)\text{ is an open set of }X_\lambda\}.\end{aligned}$$

This is clearly a topology on X and the finest for which f _λ is continuous. We just take the intersection of these topologies for all λ ∈ Λ. It also appears that a mapping f from X into a topological space Y is continuous for the inductive topology if and only if all the mappings f ∘ f _λ are continuous from X _λ to Y . Here too, why look for the finest topology in this case? This is because it lives on X which is the codomain of all f _λ . Removing sets from a topology on X can only improve the continuity status of these mappings. The situation is actually opposite to that of the projective topology, with all arrows reversed.

Let us apply this to the case of $\mathcal {D}(\Omega )$. We make use of the fact that the open set Ω has an exhaustive sequence of compact subsets , $\bigcup _{n\in {\mathbb N}}K_n=\Omega$. Let $\iota _n\colon \mathcal {D}_{K_n}(\Omega )\to \mathcal {D}(\Omega )$ be the canonical embedding simply given by inclusion. Since any compact subset of Ω is included in one of the K _n , it follows that $\mathcal {D}(\Omega )=\bigcup _n\mathcal {D}_{K_n}(\Omega )$. We equip $\mathcal {D}(\Omega )$ with the inductive topology associated with the family ι _n , after having checked that it does not depend on a specific choice of exhaustive sequence of compacts.¹⁴ The topology $\mathcal {O}_n$ associated with ι _n is a Fréchet topology, hence a locally convex topological vector space topology. Since an intersection of convex sets is convex, the inductive topology on $\mathcal {D}(\Omega )$ is also a locally convex topological vector space topology.

As a matter of fact, we also have another family of embeddings $\iota _{nm}\colon \mathcal {D}_{K_n}(\Omega )\to \mathcal {D}_{K_m}(\Omega )$ for n ≤ m that commute with the original embeddings, since the compact sets K _n are ordered under inclusion. Moreover, the topology induced on $\mathcal {D}_{K_n}(\Omega )$ by that of $\mathcal {D}_{K_m}(\Omega )$ when m ≥ n coincides with the topology of $\mathcal {D}_{K_n}(\Omega )$, because the seminorms coincide. In this case, we talk about a strict inductive limit topology, denoted by

$$\displaystyle \begin{aligned}\mathcal{D}(\Omega)=\lim_{\longrightarrow}\mathcal{D}_{K_n}(\Omega).\end{aligned}$$

This is thus the finest topology such that all embeddings ι _n are continuous. An open set U in $\mathcal {D}(\Omega )$ is a subset of $\mathcal {D}(\Omega )$ such that $U\cap \mathcal {D}_{K_n}(\Omega )$ is open in $\mathcal {D}_{K_n}(\Omega )$ for all n, which means that

$$\displaystyle \begin{aligned}U\text{ is open }\Leftrightarrow\forall\varphi\in U,\forall n\text{ such that }\mathop{\mathrm{supp}}\varphi\subset K_n; \exists p_n,\alpha_n, V_{K_n,p_n,\alpha_n}(\varphi)\subset U,\end{aligned}$$

with an obvious notation $V_{K_n,p,\alpha }(\varphi )$ for the neighborhood basis of $\mathcal {D}_{K_n}(\Omega )$ .

We note that the sequence of topologies $\mathcal {O}_n$ that are intersected, is decreasing for inclusion. In a sense, we are imposing more and more restrictions on the sets considered as n increases. Let us also note that a nonempty open set necessarily contains functions with arbitrarily large support. In fact, all $\mathcal {D}_{K}(\Omega )$ have empty interior in $\mathcal {D}(\Omega )$. Indeed, let K be a compact subset of Ω and U an open set such that $U\cap \mathcal {D}_K(\Omega )$ is nonempty, containing some function φ. Let n be such that . Since $U\cap \mathcal {D}_{K_n}(\Omega )$ is a nonempty open set of $\mathcal {D}_{K_n}(\Omega )$, it contains a neighborhood $V_{K_n,p_n,\alpha _n}(\varphi )$. Now, this neighborhood is not included in $\mathcal {D}_{K}(\Omega )$ because it contains functions whose support is strictly larger than K. To see this, add to φ a small “bump” whose support is located in K _n  ∖ K, see Fig. 1.4. Consequently, $U\not \subset \mathcal {D}_{K}(\Omega )$.

Fig. 1.4

All $\mathcal {D}_{K}(\Omega )$ have empty interior in $\mathcal {D}(\Omega )$

We also see that $\mathcal {D}_{K}(\Omega )$ is closed in $\mathcal {D}(\Omega )$. Take $\varphi \in \mathcal {D}(\Omega )\setminus \mathcal {D}_{K}(\Omega )$, and let x∉K be such that φ(x) ≠ 0. Choosing n such that $\varphi \in \mathcal {D}_{K_n}(\Omega )$, it is quite obvious that $V_{K_n,p_0,|\varphi (x)|/2}(\varphi ) \subset \mathcal {D}(\Omega )\setminus \mathcal {D}_{K}(\Omega )$ since no element of this neighborhood can vanish at point x.

The convergence of a sequence in $\mathcal {D}(\Omega )$ is actually given by Proposition 1.1. Indeed, let us take a sequence φ _k that satisfies conditions i) and ii) of the Proposition. By condition i), there exists n ₀ such that all φ _k have support in $K_{n_0}$. Condition ii) then asserts that φ _k  → φ in $\mathcal {D}_{K_{n_0}}(\Omega )$. The continuity of $\iota _{n_0}$ in turns implies that φ _k  → φ in $\mathcal {D}(\Omega )$ for the inductive limit topology.

Conversely, assume that φ _k  → φ in $\mathcal {D}(\Omega )$. Let us prove that condition i) holds. Once it is established, condition ii) is trivial. It is enough to consider the case φ = 0. Indeed, $\mathop {\mathrm {supp}}(\varphi _k-\varphi +\varphi )\subset \mathop {\mathrm {supp}}(\varphi _k-\varphi )\cup \mathop {\mathrm {supp}}(\varphi )$ which is a compact subset of Ω. Let thus U be an open set that contains 0. It follows that there exists k ₀ such that φ _k  ∈ U for all k ≥ k ₀.

We argue by contradiction with an especially well picked open set U. Let us thus assume that condition i) is not satisfied. Then for all m, there exists k _m such that $\mathop {\mathrm {supp}}\varphi _{k_m}\not \subset K_m$ which implies that there exists x _m  ∈ Ω∖ K _m with $\varphi _{k_m}(x_m)\neq 0$. Let $\ell (m)=\min \{\ell ;x_m\in K_\ell \}$. We set

$$\displaystyle \begin{aligned}p(\psi)=\sum_{m=0}^{+\infty}2\max_{x\in K_{\ell(m)}\setminus K_m}\Big|\frac{\psi(x)}{\varphi_{k_m}(x_m)}\Big|.\end{aligned}$$

We first observe that this quantity is well-defined on $\mathcal {D}(\Omega )$, because for all ψ with compact support, there is only a finite number of nonzero terms. It is in fact fairly clear that this is a seminorm on $\mathcal {D}(\Omega )$. Let us now choose

$$\displaystyle \begin{aligned}U=\{\psi\in \mathcal{D}(\Omega);p(\psi)<1\}.\end{aligned}$$

For any n, there is only a fixed finite number of nonzero terms in the sum on any $\mathcal {D}_{K_n}(\Omega )$, hence there is a constant C _n such that p ≤ C _n p ₀ on $\mathcal {D}_{K_n}(\Omega )$. It follows that p is continuous on $\mathcal {D}(\Omega )$ and that U is open for the inductive topology of $\mathcal {D}(\Omega )$. Of course 0 ∈ U, but on the other hand, $p(\varphi _{k_n})\ge 2$, which implies that $\varphi _{k_n}\notin U$, contradiction.

A few final words on the space $\mathcal {D}(\Omega )$. It is not normable because a strict inductive limit of a sequence of Montel spaces is a Montel space. In fact, its topology is not metrizable . Of course, this is a question of uncountable neighborhood bases, but we can also see it simply by using a variant of the function g of Lemma 1.2 in one dimension. Let us set

$$\displaystyle \begin{aligned}\varphi_{k,n}(x)= \begin{cases}e^{\frac k{(x-\frac 1n)^2-\frac 1{4n^2}}}&\text{for }\frac 1{2n}<x<\frac 3{2n},\\ 0&\text{otherwise,} \end{cases}\end{aligned}$$

with k > 0 and n ≥ 1. By construction, $\varphi _{k,n}\in \mathcal {D}(]0,2[)$. Moreover, for n fixed, φ _k,n → 0 in $\mathcal {D}(]0,2[)$ when k → +∞ because of the exponential term that turns up as a factor in all the derivatives. If the topology was metrizable, the usual double limit argument would enable us to find a sequence k(n) such that φ _k(n),n → 0 in $\mathcal {D}(]0,2[)$ when n → +∞. This is obviously not the case, because such a sequence cannot satisfy the support condition i) since $\mathop {\mathrm {supp}}\varphi _{k(n),n}=[\frac 1{2n},\frac 3{2n}]$.

Another amusing way of showing that $\mathcal {D}(\Omega )$ is not metrizable is to call on completeness. There is a general theory of so-called uniform structures which makes it possible to extend the concept of completeness to non metrizable spaces by replacing Cauchy sequences with more general objects called Cauchy filters, see [63]. Now it so happens that a strict inductive limit of Fréchet spaces is complete in this generalized sense.¹⁵ Since $\mathcal {D}(\Omega )=\cup _{n\in {\mathbb N}}\mathcal {D}_{K_n}(\Omega )$ and each $\mathcal {D}_{K_n}(\Omega )$ is closed with empty interior, we are thus faced with a complete space which is not a Baire space. It can then certainly not be metrizable.

Let us at last talk rapidly about the space of distributions $\mathcal {D}'(\Omega )$ , the dual space of $\mathcal {D}(\Omega )$, from the topological point of view. Dually to what was seen above, we have restriction mappings $r_n\colon \mathcal {D}^*(\Omega )\to \mathcal {D}_{K_n}^*(\Omega )$ defined by 〈r _n T, φ〉 = 〈T, ι _n φ〉 (we are taking here the algebraic duals, without continuity condition). The definition of inductive limit topology implies that T is continuous if and only if r _n T is continuous for all n, that is to say according to Proposition 1.19, the condition of Proposition 1.2. Indeed, T is continuous if and only if for any open set ω of ${\mathbb R}$, T ⁻¹(ω) is open, or again if and only if $T^{-1}(\omega )\cap \mathcal {D}_{K_n}(\Omega )$ is open in $\mathcal {D}_{K_n}(\Omega )$ for all n.

Concerning the condition of Proposition 1.3, let us consider a linear form T such that 〈T, φ _k 〉→〈T, φ〉 as soon as φ _k  → φ. In particular, this convergence holds for all sequences supported in K _n . Since $\mathcal {D}_{K_n}(\Omega )$ is metrizable, we conclude that r _n T is continuous for all n, which implies that T itself is continuous, i.e., a distribution.

We equip $\mathcal {D}'(\Omega )$ with the projective topology associated with the restriction mappings r _n and spaces $\mathcal {D}^{\prime }_{K_n}(\Omega )$ equipped with their weak-star topology. The compacts K _n are ordered by inclusion, and we then talk of a projective limit topology, which is denoted

$$\displaystyle \begin{aligned}\mathcal{D}'(\Omega)=\lim_{\longleftarrow}\mathcal{D}^{\prime}_{K_n}(\Omega).\end{aligned}$$

This is once more a locally convex topological vector space topology .

The open sets of the $\mathcal {D}'(\Omega )$ are of very little practical interest for applications to partial differential equations. They are however easy to describe since this is a projective topology. Indeed an open set must be of the form $U=\bigcup _{\lambda \in \varLambda }\bigl (\bigcap _{i=1}^{k_\lambda }r_{n_i}^{-1}(U_i)\bigr )$ where U _i is an open set of the weak-star topology of $\mathcal {D}^{\prime }_{K_{n_i}}(\Omega )$. Each open set U _i is itself of the form $V=\bigcup _{\mathrm {arbitrary}}\bigl (\bigcap _{\mathrm {finite}}\{T\in \mathcal {D}^{\prime }_{K}(\Omega );|\langle T,\varphi \rangle -a|<\varepsilon \}\bigr )$ with $\varphi \in \mathcal {D}_{K}(\Omega )$, $a\in {\mathbb R}$ and ε > 0, where we have dropped all multiple indices for legibility. Now

$$\displaystyle \begin{aligned} r_n^{-1}\bigl(\{T\in\mathcal{D}^{\prime}_{K_n}(\Omega);|\langle T,\varphi\rangle-a|<\varepsilon\}\bigr)=\{T\in\mathcal{D}'(\Omega);|\langle T,\varphi\rangle-a|<\varepsilon\}\big), \end{aligned}$$

since there is no harm in identifying a function $\varphi \in \mathcal {D}_{K_n}(\Omega )$ with its extension by 0 to Ω, which is an element of $\mathcal {D}(\Omega )$. We have seen earlier in Proposition 1.20 that a finite intersection of arbitrary unions of finite intersections of sets is itself an arbitrary union of finite intersections. Therefore, it turns out that any open set of $\mathcal {D}'(\Omega )$ is of the form $U=\bigcup _{\mathrm {arbitrary}}\bigl (\bigcap _{\mathrm {finite}}\{T\in \mathcal {D}'(\Omega );|\langle T,\varphi \rangle -a|<\varepsilon \}\bigr )$ with $\varphi \in \mathcal {D}(\Omega )$, $a\in {\mathbb R}$ and ε > 0. We thus see that the projective limit topology is also the weak-star topology on the dual space of $\mathcal {D}(\Omega )$.

On the other hand, the convergence of sequences of distributions is of paramount interest for the applications we have in mind. Due to the general properties of projective topologies, a sequence of distributions T _k converges to T if and only if r _n T _k  → r _n T in $\mathcal {D}^{\prime }_{K_n}(\Omega )$ for all n. Proposition 1.4 follows right away, given what we know about convergence in $\mathcal {D}^{\prime }_{K_n}(\Omega )$.

We notice that the topology of $\mathcal {D}'(\Omega )$ is not a metrizable topology either. Indeed, let $T_{k,n}=\frac 1k\delta ^{(n)}\in \mathcal {D}'({\mathbb R})$, where δ ⁽ⁿ⁾ is the nth derivative of the Dirac mass . For n fixed, obviously T _k,n → 0 when k → +∞. Let k(n) be any sequence of integers tending to infinity. By a theorem of Borel , there exists a function $\varphi \in \mathcal {D}({\mathbb R})$ such that φ ⁽ⁿ⁾(0) = k(n) for all n. Therefore 〈T _k(n),n, φ〉 = 190 so that T _k(n),n does not converge to 0 in the sense $\mathcal {D}'({\mathbb R})$. Now, if the topology was metrizable, we could find such a sequence k(n) for which this convergence to 0 would hold.

We have thus more or less covered all the practical properties of distributions.