Analysis Problem of the Day 53

Today’s problem appeared as Problem 10 on the Stanford Spring 2022 Analysis Qual:

Problem 5: Let $p,q \in C_c(\mathbb{T}), p>0$ be multiplication operators on the circle, and consider the operator $P = -\frac{d}{dx} p \frac{d}{dx} +q.$

a) Show that $P: H^1(\mathbb{T}) \to H^{-1}(\mathbb{T})$ is continuous and symmetric with respect to the $L^2$ inner product.

b) Show that on $H^1,$ $\langle Pu,u \rangle \geq C\|u\|_{H^1}^2-C'\|u\|_{L^2}^2$ and one may take $C'=0$ if $q=0.$

c) Show that $\|u\|_{H^1} \leq C_1(\|Pu\|_{H^{-1}}+\|u\|_{L^2})$ and $\|u\|_{H^1} \leq C_1 \|Pu\|_{H^{-1}}$ if $C'=0.$

d) Show that if $C'=0$ then $P: H^1 \to H^{-1}$ is invertible.

e) Show that $\text{ker }P$ is finite dimensional, $\text{Ran }P$ is closed and has finite codimension.

Solution: a) Since multiplication operators are symmetric and $\langle \frac{d}{dx} u,v \rangle = \langle u, -\frac{d}{dx} v \rangle$ by integration by parts, a straightforward calculation implies that $P$ is symmetric. Additionally,

$\|qu\|_{H^{-1}}=\|\langle n \rangle^{-1} \widehat{q} * \widehat{u}\|_2 \leq \|\widehat{q}*\widehat{u}\|_2 = \|qu\|_{L^2} \leq \sqrt{\|q^2\|_\infty} \|u\|_{L^2} \lesssim \| \langle n \rangle \widehat{u}\|_2 = \|u\|_{H^1}$

and

$\left\|-\frac{d}{dx} p \frac{d}{dx} u\right\|_{H^{-1}}=\|\langle n \rangle^{-1} (i n) (\widehat{p}*(-in \widehat{u}))\|_2 \lesssim \|p u'\|_{L^2} \leq \sqrt{\|p\|_\infty} \|u'\|_{L^2} \lesssim \|u\|_{H^1}.$

Thus, $P: H^1 \to H^{-1}$ is symmetric and continuous.

b) By integration by parts, note that

$\langle Pu,u \rangle= \int_0^1 p (u')^2 + qu^2 dx \geq \|p\|_\infty\|u\|_{H^1}^2 - (\|q\|_\infty +\|p\|_\infty)\|u\|_{L^2}^2,$

and if $q>0,$ one may take $C'=0$ and $0<C<\min(\min_{[0,1]} p, \min_{[0,1]} q).$

c) Note that b) implies that one may take $C'>C$ if $C' >0$ and $P+C' \geq C$ as an operator $H^1 \to L^2 \hookrightarrow H^{-1}.$ Thus, $P+C'-C-\epsilon$ is invertible, therefore bounded below, i.e. $\|(P+(C'-C-\epsilon))u\|_{H^{-1}}^2 \gtrsim \|u\|^2_{H^1}.$ In particular, since $\|u\|_{H^{-1}} \leq \|u\|_{L^2},$ we get

$\|u\|_{H^1}^2-2(C'-C-\epsilon) \langle Pu,u \rangle_{H^{-1}} \lesssim \|Pu\|_{H^{-1}}^2+\|u\|^2_{L^2}.$

Taking $\epsilon = C'-C>0$ thus yields the claim. Now, if $C'=0,$ then $P$ is a strictly positive operator from $H^1$ to $H^{-1},$ therefore bounded below, yielding the desired claim.

d) If $C'=0,$ then $P: H^1 \to H^{-1}$ is bounded below, i.e. it is injective and has closed range. Recall that $(H^{-1})^* \cong H^1.$ Then, since $P$ is symmetric, it is in fact self-adjoint from $H^1$ to $H^{-1},$ and self-adjoint operators have empty residual spectrum, i.e. their image is always dense. It follows that $P$ is invertible.

e) These properties all follow from $P: H^1 \to H^{-1}$ being Fredholm. Recall that a Fredholm operator is precisely one that is invertible modulo compact operators. Moreover, recall by Rellich-Kondrachov, the inclusion $H^1 \hookrightarrow L^2 \hookrightarrow H^{-1}$ is compact. Thus, $I: H^1 \to H^{-1}$ is compact, so it suffices to show that $P+CI$ is invertible for some $C>0.$ But by part c), $P+CI$ is bounded below, therefore injective and with closed range. Then, since $P+CI$ is self-adjoint, $\overline{\text{Ran }(P+CI)}=\text{ker}(P+CI)^\perp = H^{-1},$ so $P+CI$ is invertible, i.e. $P$ is Fredholm.

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