Nittka’s invariance criterion and Hilbert space valued parabolic equations in $L_p$

Abstract

Nittka gave an efficient criterion on a form defined on $L_2(\Omega )$ which implies that the associated semigroup is $L_p$-invariant for some given $p \in (1,\infty )$. We extend this criterion to the Hilbert space valued $L_2(\Omega ,H)$. As an application, we consider elliptic systems of purely second order. Our main result shows that the induced semigroup is $L_p$-contractive for all $p \in [p_-,p_+]$ for some $1< p_-< 2< p_+ < \infty$.

Appendices

A. The derivative of a truncation

Let $\Omega \subset \mathbb {R}^d$ be an open set. Let H be a Hilbert space. The principal aim in this section is to prove the following chain rule.

Proposition A.1

Let $\alpha > 0$ and $M > 0$. Let $u \in H^1(\Omega ,H)$. Define $v = (\Vert u\Vert _H^\alpha \wedge M) \, u$. Then $v \in H^1(\Omega ,H)$ and

$$\begin{aligned} \partial _k v = \alpha \, \mathbbm {1}_{[\Vert u\Vert _H^\alpha < M]} \, \Vert u\Vert _H^\alpha \, {\textrm{Re}}\,({\textrm{sgn}}\,u, \partial _k u)_H \, {\textrm{sgn}}\,u + (\Vert u\Vert _H^{\alpha } \wedge M) \, \partial _k u \end{aligned}$$

for all $k \in \{ 1,\ldots ,d \}$.

The proof involves some work. We use the following approximation by smooth functions.

Lemma A.2

The space $C^\infty (\Omega ,H) \cap H^1(\Omega ,H)$ is dense in $H^1(\Omega ,H)$.

Proof

This follows as in the scalar case in [1, Theorem 3.17]. $\square$

For an approximation argument, the next lemma is useful.

Lemma A.3

For all $n \in \mathbb {N}$, let $u_n \in H^1(\Omega ,H)$. Let $u,g_1,\ldots ,g_d \in L_2(\Omega ,H)$. Suppose that $\lim u_n = u$ in $L_2(\Omega ,H)$ and $\lim \partial _k u_n = g_k$ in $L_2(\Omega ,H)$ for all $k \in \{ 1,\ldots ,d \}$. Then $u \in H^1(\Omega ,H)$ and $\partial _k u = g_k$ for all $k \in \{ 1,\ldots ,d \}$.

Proof

Let $\varphi \in C_c^\infty (\Omega )$. Let $k \in \{ 1,\ldots ,d \}$. Then $- \int _\Omega u_n \, \partial _k \varphi = \int _\Omega (\partial _k u_n) \, \varphi$ for all $n \in \mathbb {N}$. Then the lemma follows by taking the limit $n \rightarrow \infty$. $\square$

For the proof of Proposition A.1, we shall approximate the function $t \mapsto t^\alpha \wedge M$ with smooth functions. The next technical lemma gives sufficient conditions in order to apply a chain rule. Note that we do not require that $f'$ is bounded.

Lemma A.4

Let $f \in C^1(0,\infty )$. Suppose that f is bounded, $\lim _{t \downarrow 0} f(t) = 0$, $\lim _{t \downarrow 0} t \, f'(t) = 0$, and $\sup _{t \in (0,\infty )} t \, |f'(t)| < \infty$. Let $u \in H^1(\Omega ,H)$. Define $v = f(\Vert u\Vert _H) \, u$. Then $v \in H^1(\Omega ,H)$ and

$$\begin{aligned} \partial _k v = \left\{ \begin{array}{ll} \Vert u\Vert _H \, f'(\Vert u\Vert _H) \, {\textrm{Re}}\,({\textrm{sgn}}\,u, \partial _k u)_H \, {\textrm{sgn}}\,u + f(\Vert u\Vert _H) \, \partial _k u &{} \text{ on } [u \ne 0], \\ 0 &{} \text{ on } [u = 0], \end{array} \right. \nonumber \\ \end{aligned}$$

(5)

for all $k \in \{ 1,\ldots ,d \}$.

Proof

Let $\varepsilon > 0$. Define $v_\varepsilon = f(\sqrt{\Vert u\Vert _H^2 + \varepsilon }) \, u$. If $u \in C^1(\Omega ,H)$, then $v_\varepsilon \in C^1(\Omega ,H)$ and

$$\begin{aligned} \partial _k v_\varepsilon= & {} \sqrt{\Vert u\Vert _H^2 + \varepsilon } \, f'(\sqrt{\Vert u\Vert _H^2 + \varepsilon }) \, \frac{{\textrm{Re}}\,(u, \partial _k u)_H}{\sqrt{\Vert u\Vert _H^2 + \varepsilon }} \, \frac{1}{\sqrt{\Vert u\Vert _H^2 + \varepsilon }} \, u\nonumber \\{} & {} + f(\sqrt{\Vert u\Vert _H^2 + \varepsilon }) \, \partial _k u \end{aligned}$$

(6)

for all $k \in \{ 1,\ldots ,d \}$. Then, by Lemmas A.2 and A.3, this extends to all $u \in H^1(\Omega ,H)$ and (6) is valid. Finally choose $\varepsilon = \frac{1}{n}$, take the limit $n \rightarrow \infty$, and use again Lemma A.3. $\square$

Now we are able to prove Proposition A.1.

Proof of Proposition A.1

For all $n \in \mathbb {N}$, define $f_n,f :(0,\infty ) \rightarrow \mathbb {R}$ by

$$\begin{aligned} f(t)= & {} t^\alpha \wedge M, \\ f_n(t)= & {} {\textstyle \frac{1}{2}} \, ( t^\alpha + \sqrt{M^2 + n^{-1}} - \sqrt{|t^\alpha - M|^2 + n^{-1}} ). \end{aligned}$$

Then $\lim f_n(t) = f(t)$ for all $t \in (0,\infty )$. Also $\lim _{t \downarrow 0} f_n(t) = 0$ for all $n \in \mathbb {N}$. Let $n \in \mathbb {N}$. Then $f_n \in C^1(0,\infty )$ and

$$\begin{aligned} f_n'(t) = {\textstyle \frac{1}{2}} \, \alpha \, t^{\alpha - 1} \Big ( 1 - \frac{t^\alpha - M}{\sqrt{(t^\alpha - M)^2 + n^{-1}}} \Big ) \end{aligned}$$

for all $t \in (0,\infty )$. In particular, $f_n$ is increasing. Moreover, $\lim _{t \downarrow 0} t \, f_n'(t) = 0$. In addition, $\lim _{n \rightarrow \infty } f_n'(t) = \alpha \, t^{\alpha - 1}$ if $t^\alpha \le M$ and $\lim _{n \rightarrow \infty } f_n'(t) = 0$ if $t^\alpha > M$.

Let $n \in \mathbb {N}$ and $t \in (0,\infty )$. If $t^\alpha \le M$, then

$$\begin{aligned} 0 \le t \, f_n'(t) = {\textstyle \frac{1}{2}} \, \alpha \, t^\alpha \Big ( 1 + \frac{M - t^\alpha }{\sqrt{(t^\alpha - M)^2 + n^{-1}}} \Big ) \le \alpha \, M. \end{aligned}$$

Alternatively, if $t^\alpha > M$, then

$$\begin{aligned} 0\le & {} t \, f_n'(t) = {\textstyle \frac{1}{2}} \, \alpha \, \frac{t^\alpha - M + M}{\sqrt{(t^\alpha - M)^2 + n^{-1}}} \Big ( \sqrt{(t^\alpha - M)^2 + n^{-1}} - \sqrt{(t^\alpha - M)^2} \Big ) \\\le & {} {\textstyle \frac{1}{2}} \, \alpha \Big ( 1 + \frac{M}{\sqrt{n^{-1}}} \Big ) \sqrt{n^{-1}} \le {\textstyle \frac{1}{2}} \, \alpha \, (1 + M). \end{aligned}$$

So

$$\begin{aligned} \sup _{n \in \mathbb {N}} \sup _{t \in (0,\infty )} t \, |f_n'(t)| \le \alpha \, (M+1). \end{aligned}$$

(7)

If $n \in \mathbb {N}$ and $t \in (0,\infty )$, then

$$\begin{aligned} 0\le & {} f_n(t) \le {\textstyle \frac{1}{2}} \, (t^\alpha + M+1 - \sqrt{|t^\alpha - M|^2 + n^{-1}} )\\\le & {} {\textstyle \frac{1}{2}} \, (t^\alpha + M+1 - |t^\alpha - M| ) = {\textstyle \frac{1}{2}} + f(t) \le {\textstyle \frac{1}{2}} + M. \end{aligned}$$

So $f_n$ is bounded and even

$$\begin{aligned} \sup _{n \in \mathbb {N}} \sup _{t \in (0,\infty )} |f_n(t)| \le {\textstyle \frac{1}{2}} + M. \end{aligned}$$

(8)

Hence all conditions of Lemma A.4 are satisfied for all the $f_n$.

Let $u \in H^1(\Omega ,H)$. For all $n \in \mathbb {N}$, define $v_n = f_n(u) \, u$. Then $v_n \in H^1(\Omega ,H)$ with derivatives given by (5) and f replaced by $f_n$. The Lebesgue dominated convergence theorem and the uniform bounds (8) and (7) imply that $\lim v_n = v$ and $\lim \partial _k v_n = \partial _k v$ in $L_2(\Omega ,H)$ for all $k \in \{ 1,\ldots ,d \}$. Then the proposition follows from Lemma A.3. $\square$

Almost the same arguments show that the norm of an $H^1(\Omega ,H)$-function is in the Sobolev space.

Lemma A.5

Let $u \in H^1(\Omega ,H)$. Then $\Vert u\Vert _H \in H^1(\Omega )$ and furthermore $\partial _k \Vert u\Vert _H = {\textrm{Re}}\,({\textrm{sgn}}\,u, \partial _k u)_H$ for all $k \in \{ 1,\ldots ,d \}$.

Proof

Let $\varepsilon > 0$. For all $u \in H^1(\Omega ,H)$, define $u_\varepsilon :\Omega \rightarrow H$ by $u_\varepsilon = \sqrt{\Vert u\Vert _H^2 + \varepsilon }$. Let $\varphi \in C_c^\infty (\Omega )$ and $k \in \{ 1,\ldots ,d \}$. If $u \in C^\infty (\Omega ,H) \cap H^1(\Omega ,H)$, then $u_\varepsilon \in C^\infty (\Omega ,H)$ with classical partial derivative $\partial _k u_\varepsilon = \frac{{\textrm{Re}}\,(u, \partial _k u)_H}{u_\varepsilon }$. Hence

$$\begin{aligned} - \int \limits _\Omega u_\varepsilon \, \partial _k \varphi = \int \limits _\Omega \frac{{\textrm{Re}}\,(u, \partial _k u)_H}{u_\varepsilon } \, \varphi . \end{aligned}$$

(9)

Using approximation and Lemma A.2, it follows that (9) is valid for all $u \in H^1(\Omega ,H)$. Finally choose $\varepsilon = \frac{1}{n}$ and take the limit $n \rightarrow \infty$. $\square$

B. Strict convexity of $L_p(\Omega ,H)$

As before, let $(\Omega ,{{\mathcal {B}}},\mu )$ be a $\sigma$-finite measure space and H a Hilbert space. In order to make this paper more self-contained, we give a direct proof of the following theorem. At the end of this section, we give information on more general results.

Theorem B.1

Let $p \in (1,\infty )$ and $u,v \in L_p(\Omega ,H)$ with $\Vert u\Vert _p = \Vert v\Vert _p = 1$. If $\Vert u+v\Vert _p = 2$, then $u = v$.

For the proof of Theorem B.1, we use three lemmas.

Lemma B.2

Let $\xi ,\eta \in H$ and suppose that $\Vert \xi + \eta \Vert _H = \Vert \xi \Vert _H + \Vert \eta \Vert _H$. If $\eta \ne 0$, then there is a $\lambda \in [0,\infty )$ such that $\xi = \lambda \, \eta$.

Proof

The equality implies that ${\textrm{Re}}\,(\xi ,\eta )_H = \Vert \xi \Vert _H \, \Vert \eta \Vert _H$. This gives equality in the Cauchy–Schwarz inequality. Hence there is a $\lambda \in \mathbb {C}$ such that $\xi = \lambda \, \eta$. Using again the equality, one deduces that $|1 + \lambda | = |\lambda | + 1$ and therefore $\lambda \in [0,\infty )$. $\square$

Let $p,q \in (1,\infty )$ and suppose that $\frac{1}{p} + \frac{1}{q} = 1$.

Lemma B.3

Let $a,b \in [0,\infty )$. Then $a \, b \le \frac{1}{p} \, a^p + \frac{1}{q} \, b^q$ and the equality holds if and only if $a^p = b^q$.

Proof

This follows from the concavity of the logarithm. $\square$

Lemma B.4

Let $f \in L_p(\Omega )$ and $g \in L_q(\Omega )$ with $f \ne 0$ and $g \ne 0$. Suppose that $\int _\Omega |f| \, |g| = \Vert f\Vert _{L_p(\Omega )} \, \Vert g\Vert _{L_q(\Omega )}$. Then there exists a $\lambda > 0$ such that $|f|^p = \lambda \, |g|^q$ almost everywhere.

Proof

We may assume that $\Vert f\Vert _{L_p(\Omega )} = 1 = \Vert g\Vert _{L_q(\Omega )}$. Then

$$\begin{aligned} 1 = \int \limits _\Omega |f| \, |g| \le \int \limits _\Omega {\textstyle \frac{1}{p}} \, |f|^p + {\textstyle \frac{1}{q}} \, |g|^q = {\textstyle \frac{1}{p}} \, \Vert f\Vert _{L_p(\Omega )} + {\textstyle \frac{1}{q}} \, \Vert g\Vert _{L_q(\Omega )} = 1. \end{aligned}$$

Hence $|f| \, |g| = {\textstyle \frac{1}{p}} \, |f|^p + \frac{1}{q} \, |g|^q$ almost everywhere and the lemma follows from Lemma B.3. $\square$

Proof of Theorem B.1

Using the triangle inequality on H, twice the Hölder inequality on $L_p(\Omega )$, and the assumption $\Vert u+v\Vert _p = \Vert u\Vert _p + \Vert v\Vert _p$, one obtains

$$\begin{aligned} \Vert u + v\Vert _p^p= & {} \int \limits _\Omega \Vert u+v\Vert _H \, \Vert u+v\Vert _H^{p-1}\\\le & {} \int \limits _\Omega \Vert u\Vert _H \, \Vert u+v\Vert _H^{p-1} + \int \limits _\Omega \Vert v\Vert _H \, \Vert u+v\Vert _H^{p-1} \\\le & {} \Big ( \int \limits _\Omega \Vert u\Vert _H^p \Big )^{1/p} \Big ( \int \limits _\Omega \Vert u+v\Vert _H^{(p-1) q} \Big )^{1/q}\\{} & {} + \Big ( \int \limits _\Omega \Vert v\Vert _H^p \Big )^{1/p} \Big ( \int \limits _\Omega \Vert u+v\Vert _H^{(p-1) q} \Big )^{1/q} \\= & {} (\Vert u\Vert _p + \Vert v\Vert _p) \, \Vert u+v\Vert _p^{p/q} = \Vert u+v\Vert _p^{\frac{p}{q} + 1} = \Vert u+v\Vert _p^p. \end{aligned}$$

Hence all three inequalities are equalities. The first gives that there is a null-set $N_1 \subset \Omega$ such that $\Vert u+v\Vert _H(x) = \Vert u\Vert _H(x) + \Vert v\Vert _H(x)$ for all $x \in \Omega {\setminus } N_1$ such that $\Vert u+v\Vert _H(x) \ne 0$. Recall that $\Vert u\Vert _p = 1$, so $\Vert u\Vert _H \ne 0 \in L_p(\Omega )$. Similarly $\Vert u+v\Vert _H \ne 0 \in L_p(\Omega )$ and therefore $\Vert u+v\Vert _H^{p-1} \ne 0 \in L_q(\Omega )$. Hence the equality in the first Hölder inequality together with Lemma B.4 gives that there are $\alpha > 0$ and a null-set $N_2 \subset \Omega$ such that $\Vert u\Vert _H^p(x) = \alpha \, \Vert u+v\Vert _H^{(p-1)q}(x)$ for all $x \in \Omega {\setminus } N_2$. Similarly there are $\beta > 0$ and a null-set $N_3 \subset \Omega$ such that $\Vert v\Vert _H^p(x) = \beta \, \Vert u+v\Vert _H^{(p-1)q}(x)$ for all $x \in \Omega {\setminus } N_3$. Hence $\Vert u\Vert _H^p = \gamma \, \Vert v\Vert _H^p$ on $\Omega \setminus (N_2 \cup N_3)$, where $\gamma = \frac{\alpha }{\beta }$. Since $\Vert u\Vert _p = \Vert v\Vert _p = 1$, one deduces that $\gamma = 1$.

Now let $x \in \Omega \setminus (N_1 \cup N_2 \cup N_3)$. If $\Vert u+v\Vert _H(x) = 0$, then subsequently $\Vert u\Vert _H^p(x) = \alpha \, \Vert u+v\Vert _H^{(p-1)q}(x) = 0$ and $u(x) = 0$. Similarly $v(x) = 0$ and therefore $u(x) = v(x)$. Alternatively, if $\Vert u+v\Vert _H(x) \ne 0$, then $v(x) \ne 0$ since $x \not \in N_3$. Moreover, $\Vert u(x) + v(x)\Vert _H = \Vert u(x)\Vert _H + \Vert v(x)\Vert _H$ and Lemma B.2 implies that there is a $\lambda \in [0,\infty )$ such that $u(x) = \lambda \, v(x)$. But $\Vert u(x)\Vert _H^p = \Vert v(x)\Vert _H^p$ and hence $u(x) = v(x)$. Therefore $u = v$ almost everywhere. $\square$

With a small modification, one can prove that $L_p(\Omega ,E)$ is strictly convex if E is strictly convex and $p \in (1,\infty )$. In fact, a stronger result than Theorem B.1 is known. The space $L_p(\Omega ,E)$ is uniformly convex if E is uniformly convex and $p \in (1,\infty )$. See [10] and the references therein.