A Note on Reflected Dirichlet Forms

Abstract

In this paper we give an algebraic construction of the (active) reflected Dirichlet form. We prove that it is the maximal Silverstein extension whenever the given form does not possess a killing part and we prove that Dirichlet forms need not have a maximal Silverstein extension if a killing is present. For regular Dirichlet forms we provide an alternative construction of the reflected process on a compactification (minus one point) of the underlying space.

Appendices

Appendix A: Closed Forms on Metrizable Topological Vector Spaces

In this section we provide a short introduction to quadratic forms on metrizable topological vector spaces. All of the results presented here are special cases of the theory developed in [31, Chapter 1], which treats quadratic forms on general topological vector spaces. Since on metrizable topological vector spaces many of the arguments in [31] simplify substantially, we chose to include proofs for the convenience of the reader.

Note that all the results of this section are well known for quadratic forms on L²(m). However, their classical proofs use that L²(m) is locally convex in an essential way and in the main text we apply the theory to forms on L⁰(m), which is not locally convex in general.

Let V be a real vector space. We call q : V → [0, ∞] a quadratic form on V if it satisfies

• q(λf) = |λ|²q(f) for all $\lambda \in {\mathbb {R}}$, f ∈ V,

• q(f + g) + q(f − g) = 2q(f) + 2q(g) for all f, g ∈ V.

Here we use the conventions ∞⋅ 0 = 0, ∞⋅ x = ∞ for x ∈ (0, ∞] and y + ∞ = ∞ for y ∈ [0, ∞]. The domain of a quadratic form q is D(q) = {f ∈ V ∣q(f) < ∞} and its kernel is ker q = {f ∈ V ∣q(f) = 0}. Both sets are subspaces of V. For checking whether or not a functional is a quadratic form, one does not need to verify identities but certain inequalities on its domain.

Lemma A.1

Forq : V → [0, ∞] the following assertions are equivalent.

• (i) q is a quadratic form.

• (ii) For all$\lambda \in {\mathbb {R}}$andallf, g ∈ Vwithq(f), q(g) < ∞theinequalities

  $$2q(f) + 2q(g) \leq q(f + g) + q(f-g) \text{ and } q(\lambda f) \leq |\lambda|^{2} q(f) $$

  hold.

By a theorem of Jordan and von Neumann, see [17], any quadratic form induces a bilinear form on its domain via polarization, i.e., the mapping

$$q:D(q) \times D(q) \to {\mathbb{R}}, (f,g) \mapsto q(f,g) := \frac{1}{4}\left( q(f+g) - q(f-g)\right) $$

is bilinear. We abuse notation and write q both for the quadratic form on V and the induced bilinear form on D(q). In this sense, we have q(f) = q(f, f) for f ∈ D(q). Values of the form q(f) = q(f, f) are called on-diagonal values ofq and values of the form q(f, g) for different f, g ∈ D(q) are called off-diagonal values ofq. It is important to note that q (as a bilinear form) satisfies the Cauchy-Schwarz inequality and q^1/2 (as the square root of a quadratic form) is a seminorm on D(q).

A quadratic form $\tilde {q}$ is called an extension of the quadratic form q if $D(q) \subseteq D(\tilde {q})$ and $q(f) = \tilde q(f)$ for all f ∈ D(q). There is a natural partial order on the cone of all quadratic forms on V. We say that two quadratic forms $q,\tilde q$ on V satisfy $q \leq \tilde q$ if $q(f) \geq \tilde q(f)$ for all f ∈ V. Large forms in terms of this relation are the ones with large domains. Indeed, we have $q \leq \tilde q$ if and only if $D(q) \subseteq D(\tilde q)$ and $q(f) \geq \tilde q(f)$ for all f ∈ D(q).

A (real) metrizable topological vector space is a real vector space V equipped with a balanced translation invariant metric ρ : V × V → [0, ∞), i.e., a metric such that for all f, g, h ∈ V and $\lambda \in {\mathbb {R}}$ with |λ|≤ 1 we have

$$\rho(\lambda f,0) \leq \rho(f,0) \text{ and } \rho(f + h,g+h) = \rho(f,g). $$

The vector space operations are continuous with respect to such a metric. For a sequence (f_n) in (V, ρ) that converges to f ∈ V with respect to ρ we write $f_{n} \overset {\rho }{\to } f$ for short.

Remark 1

In the text we shall mainly consider V = L²(m) or V = L⁰(m) for some σ-finite measure m on X. In the latter case a translation invariant metric is given as follows. For an ascending sequence of sets of finite measure (F_n) with ∪_nF_n = X we let

$$d(f,g) := \sum\limits_{n = 1}^{\infty} \frac{1}{2^{n} m(F_{n})} {\int}_{F_{n}} |f-g| \wedge 1 d m. $$

It is a translation invariant and balanced metric and it induces the topology of local convergence in measure. Moreover, the metric space (L⁰(m), d) is complete.

In what follows we fix a metrizable topological vector space (V, ρ). The following lemma is nontrivial because balls with respect to ρ need not be convex.

Lemma A.2 (Convergent Cesàro means)

Let (f_n) be a convergent sequencein (V, ρ) with limit f. Then thereexists a subsequence$(f_{n_{k}})$such thatfor all of its subsequences$(f_{n_{k_{l}}})$wehave

$$\underset{N \to \infty}{\lim} \frac{1}{N} \sum\limits_{l = 1}^{N} f_{n_{k_{l}}} = f. $$

Proof

Since ρ is translation invariant, we may assume f = 0. We choose a subsequence $(f_{n_{k}})$ such that

$$\sum\limits_{k = 1}^{\infty} \rho(f_{n_{k}},0) < \infty. $$

For an arbitrary subsubsequence $(f_{n_{k_{l}}})$ the translation invariance of ρ implies

$$\rho\left( \frac{1}{N} \sum\limits_{l = 1}^{N} f_{n_{k_{l}}}, 0\right) \leq \rho\left( \frac{1}{N} \sum\limits_{l = 1}^{M} f_{n_{k_{l}}},0\right) + \sum\limits_{l = M + 1}^{N}\rho\left( \frac{1}{N}f_{n_{k_{l}}},0\right). $$

From this inequality and the balancedness of ρ we infer

$$\rho\left( \frac{1}{N} \sum\limits_{l = 1}^{N} f_{n_{k_{l}}}, 0\right) \leq \rho\left( \frac{1}{N} \sum\limits_{l = 1}^{M} f_{n_{k_{l}}},0\right) + \sum\limits_{l = M + 1}^{\infty} \rho\left( f_{n_{k_{l}}},0\right). $$

Choosing first M and then N large enough and using the continuity of the multiplication with scalars at 0 finishes the proof. □

We call a quadratic form q on a metrizable topological vector space (V, ρ) closed if it is lower semicontinuous with respect to ρ-convergence, i.e., if $f_{n} \overset {\rho }{\to } f$ implies

$$q(f) \leq \underset{n \to \infty}{\liminf} q(f_{n}). $$

A quadratic form is called closable if it possesses a closed extension. For determining whether or not a quadratic form possesses a closed extension the following lemma is useful.

Lemma A.3

A quadratic form qon (V, ρ) is closable if and only if it is lower semicontinuous on its domain, i.e., if for all sequences(f_n) inD(q) andf ∈ D(q) the convergence$f_{n} \overset {\rho }{\to } f$implies

$$q(f) \leq \underset{n\to \infty}{\liminf} q(f_{n}). $$

In this case, it possesses a smallest closed extension$\bar q:V \to [0,\infty ]$that is given by

$$\bar q (f) = \left\{\begin{array}{ll} \underset{n \to \infty}{\lim} q(f_{n}) &\text{if } (f_{n}) \text{ is }q\text{-Cauchy with } f_{n} \overset{\rho}{\to} f,\\ \infty &\text{if there exists no such sequence}. \end{array}\right. $$

Proof

The necessity of lower semicontinuity on the domain for the existence of a closed extension is clear and so it suffices to show that this condition is sufficient. Recall that a function F : V → (−∞, ∞] is lower semicontinuous if and only if its epigraph

$$\text{epi} F = \{(f,t) \in V \times {\mathbb{R}} \mid F(f) \leq t\} $$

is closed in the product space $V \times {\mathbb {R}}$ (indeed this is true for functions on any metric space).

Let $\overline {\text {epi} q}$ the closure of epiq in $V \times {\mathbb {R}}$. We define $\tilde q:V \to [0,\infty ]$ by

$$\tilde{q}(f) := \inf \{t \mid (f,t) \in \overline{\text{epi} q}\}, $$

where we use the convention inf ∅ = ∞. It is readily verified that $\text {epi} \tilde q = \overline {\text {epi} q}$ and so $\tilde q$ is lower semicontinuous. The lower semicontinuity of q on its domain implies that $\tilde q (f) = q(f)$ for f ∈ D(q) and that the domain of any closed extension of q needs to contain the epigraph of $\tilde q$.

Let $\bar q$ as in the statement of the lemma. The lower semicontinuity of q on its domain shows that $\bar q$ is well defined. Indeed, if (f_n), (g_n) are q-Cauchy sequences in D(q) with $f_{n} \overset {\rho }{\to } f$, $g_{n} \overset {\rho }{\to } f$, then (f_n − g_n) is a q-Cauchy sequence with $f_{n} - g_{n} \overset {\rho }{\to } 0$. Since f_n − g_n ∈ D(q) and q is lower semicontinuous on its domain, we obtain

$$|q(f_{n})^{1/2} - q(g_{n})^{1/2}| \leq q(f_{n} - g_{n})^{1/2} \leq \underset{m \to \infty}{\liminf} q(f_{n} - g_{n} - f_{m} + g_{m})^{1/2}, $$

and therefore

$$\underset{n\to \infty}{\lim} q(f_{n}) = \underset{n\to \infty}{\lim} q(g_{n}). $$

To finish the proof it now suffices to show $\bar q = \tilde q$, which implies that $\bar q$ is lower semicontinuous. That $\bar q$ is a quadratic form is immediate from the definition of $\bar q$ and its minimality follows from the minimality of $\tilde q$.

Let f ∈ V. We first prove $\tilde q(f) \leq \bar q(f)$. If $\bar q(f) = \infty$ there is nothing to show. Assume that there is a q-Cauchy sequence (f_n) in D(q) with $f_{n} \overset {\rho }{\to } f$. The lower semicontinuity of $\tilde q$ and that it coincides with q on D(q) implies

$$\tilde q (f) \leq \underset{n\to \infty}{\liminf} \tilde q(f_{n}) = \underset{n\to \infty}{\liminf} q(f_{n}) = \bar q (f). $$

For proving the opposite inequality we can assume $\tilde q(f) < \infty$. Since $(f,\tilde q(f)) \in \text {epi} \tilde q = \overline {\text {epi} q}$, the definition of $\overline {\text {epi} q}$ yields that there exists a sequence (f_n, t_n) ∈epiq with $f_{n} \overset {\rho }{\to } f$ and $t_{n} \to \tilde q(f)$. In particular, q(f_n) ≤ t_n and so the sequence (f_n) is q-bounded. By Lemma A.2 and by the Banach-Saks theorem, see e.g. [7, Theorem A.4.1], we can pass to a suitable subsequence and additionally assume that the sequence of Cesàro means

$$g_{N} := \frac{1}{N}\sum\limits_{n = 1}^{N} f_{n} $$

is q-Cauchy and satisfies $g_{N} \overset {\rho }{\to } f$. Using the definition of $\bar q$ we obtain

$$\bar q(f)^{1/2} = \underset{N \to \infty}{\lim} q(g_{N})^{1/2} \leq \underset{N \to \infty}{\liminf} \frac{1}{N} \sum\limits_{n = 1}^{N} q(f_{n})^{1/2} = \tilde q(f)^{1/2}. $$

This finishes the proof. □

When the underlying topological vector space (V, ρ) is complete, we can give another useful characterization of closed forms. To this end, we consider

$$\rho_{q} :D(q) \times D(q) \to [0,\infty), \rho_{q}(f,g):= \rho(f,g) + q(f-g)^{1/2}. $$

Since q^1/2 is a seminorm, it is a translation invariant balanced metric on D(q). The topology generated by ρ_q is called the form topology and we call a Cauchy sequence with respect to ρ_q a Cauchy sequence with respect to the form topology. The latter naming is a bit imprecise as there is no Cauchyness with respect to topologies. However, since the form topology is a vector space topology, it has a canonical uniform structure which coincides with the uniform structure induced by the metric ρ_q.

Lemma A.4

Let (V, ρ) be complete. The following assertions are equivalent.

• (i) q is a closed quadratic form on (V, ρ).

• (ii) (D(q), ρ_q) is a complete metric space.

Proof

(i) ⇒ (ii): Let (f_n) in D(q) Cauchy with respect to ρ_q. Since (V, ρ) is complete it has a ρ-limit f ∈ V. The lower semicontinuity of q implies f ∈ D(q) and

$$q(f - f_{n}) \leq \underset{m \to \infty}{\liminf} q(f_{m} - f_{n}). $$

This shows f_n → f with respect to ρ_q and the completeness of is proven.

(ii) ⇒ (i): Let (f_n) a sequence in V and f ∈ V with $f_{n} \overset {\rho }{\to } f$. We prove $q(f) \leq \liminf _{n} q(f_{n})$. After passing to a suitable subsequence we can assume

$$\underset{n \to \infty}{\liminf} q(f_{n}) = \underset{n\to \infty}{\lim} q(f_{n}) < \infty. $$

The Banach-Saks theorem and Lemma A.2 imply that after passing to a further subsequence we can assume that the sequence of Cesàro means

$$g_{N} := \frac{1}{N}\sum\limits_{n = 1}^{N} f_{n}$$

is q-Cauchy and satisfies $g_{N} \overset {\rho }{\to } f$, as N →∞. The completeness of (D(q), ρ_q) yields g_N → f with respect to q and we obtain

$$q(f)^{1/2} = \underset{N \to \infty}{\lim} q(g_{N})^{1/2} \leq \underset{N\to \infty}{\liminf} \sum\limits_{n = 1}^{N} q(f_{n})^{1/2} = \underset{n\to \infty}{\lim} q(f_{n})^{1/2}.$$

This finishes the proof. □

The following lemma on weakly convergent sequences is quite useful, for an L²-version, see e.g. [24, Lemma I.2.12].

Lemma A.5

Letq be a closed quadratic form on (V, ρ).Let (f_n) be a q-boundedsequence inD(q) and letf ∈ Vwith$f_{n} \overset {\rho }{\to } f$.Thenf ∈ D(q) and (f_n) converges q-weakly f. If, additionally,

$$\underset{n\to \infty}{\limsup} q(f_{n}) \leq q(f),$$

thenf_n → fwith respect toq.

Proof

We first show the statement on weak convergence. The lower semicontinuity of q and the q-boundedness of (f_n) imply f ∈ D(q). Hence, by considering the sequence (f − f_n) instead of (f_n), we can assume f = 0.

The boundedness of (f_n) implies that for each g ∈ D(q) we have

$$- \infty < \underset{n\to \infty}{\liminf} q(f_{n},g) \leq \underset{n\to \infty}{\limsup} q(f_{n},g) < \infty. $$

Let M ≥ 0 such that q(f_n) ≤ M for each n. Since the vector space operations are continuous with respect to ρ, for α > 0 and g ∈ D(q) we have $g-\alpha f_{n} \overset {\rho }{\to } g$. The lower semicontinuity of q yields

$$\begin{array}{@{}rcl@{}} q(g) &\leq& \underset{n \to \infty}{\liminf} q(g - \alpha f_{n})\\ &=& \underset{n \to \infty}{\liminf} \left( q(g) - 2\alpha q(f_{n},g) + \alpha^{2} q(f_{n}) \right) \\ &\leq& \underset{n \to \infty}{\liminf} \left( q(g) - 2\alpha q(f_{n},g) + \alpha^{2} M \right) \\ &=&q(g) - 2\alpha \underset{n \to \infty}{\limsup} q(f_{n},g) + \alpha^{2} M. \end{array} $$

Hence, for all α > 0 we obtain

$$2\underset{n \to \infty}{\limsup} q(f_{n},g) \leq \alpha M,$$

which implies

$$\underset{n \to \infty}{\limsup} q(f_{n},g) \leq 0.$$

Since g ∈ D(q) was arbitrary, we also have $\limsup _{n \to \infty } q(f_{n},-g) \leq 0$ and conclude

$$0 \leq \liminf_{n \to \infty} q(f_{n},g) \leq \limsup_{n \to \infty} q(f_{n},g) \leq 0.$$

This shows weak convergence. The strong convergence under the additional condition $\limsup _{n\to \infty } q(f_{n}) \leq q(f)$ follows from the weak convergence and the identity

$$q(f-f_{n}) = q(f) + q(f_{n}) - 2 q(f,f_{n}).$$

This finishes the proof. □

Appendix B: A Lemma on Monotone Forms

We call a quadratic form q : L⁰(m) → [0, ∞] nonnegative definite if for f, g ∈ D(q) the inequality fg ≥ 0 implies q(f, g) ≥ 0. We call it monotone if for f, g ∈ D(q) the inequality |f|≤|g| implies q(g) ≤ q(f). The following lemma shows that monotone and nonnegative definite quadratic forms coincide when their domain is a lattice in the sense of the natural order on L⁰(m), i.e., if f, g ∈ D(q) implies f ∧ g ∈ D(q) and f ∨ g ∈ D(q).

Lemma B.1

Letq be a quadratic form onL⁰(m) such thatD(q) is a lattice. The following assertions are equivalent.

• (i) q is monotone.

• (ii) q is nonnegative definite.

Proof

(i) ⇒ (ii): For f, g ∈ D(q) with fg ≥ 0 we have |f + g|≥|f − g|. The monotonicity of q implies

$$q(f) + q(g) - 2q(f,g) = q(f-g) \leq q(f+g) = q(f) + q(g) + 2q(f,g)$$

proving that it is nonnegative definite.

(ii) ⇒ (i): We first prove that q(f) = q(|f|) for f ∈ D(q). Since D(q) is a lattice, f₊ = f ∨ 0 and f₋ = (−f) ∨ 0 belong to D(q). We obtain

$$q(f) = q(f_{+}) + q(f_{-}) - 2q(f_{+},f_{-}) \text{ and } q(|f|) = q(f_{+}) + q(f_{-}) + 2q(f_{+},f_{-}). $$

Since f₊f₋ = 0, the positive definiteness of q yields q(f₊, f₋) = 0, proving q(f) = q(|f|).

Now, let f, g ∈ D(q) with |f|≤|g| be given. Using the inequalities |f|(|g|−|f|) ≥ 0 and |g|(|g|−|f|) ≥ 0, the positive definiteness of q implies

$$q(f) = q(|f|) \leq q(|g|,|f|) \leq q(|g|) = q(g).$$

This finishes the proof. □