6
$\begingroup$

I have the following question:

Let $f$ be a continuous function in an open, bounded, smooth domain $\Omega$ in $\mathbb{R}^{n}$ such that $| \int_{\Omega} fg|\leq 1$ for any $g\in C_{0}^{\infty}(\Omega)$, $\|g\|_{L^{2}(\Omega)}=1$. Here the measure and the integral are w.r.t. Lebesgue measure. And $C_{0}^{\infty}(\Omega)$ is the set of smooth functions of compact support in $\Omega$.

Can we conclude that $f\in L^{2}(\Omega)$?

Thanks for any hint.

$\endgroup$
6
  • $\begingroup$ What are we assuming about $f$ to make $\int_{\Omega} fg$ an a priori well-defined quantity? If $f$ is measurable and not in $L^1_{loc}$ already, the expression need not be defined. $\endgroup$ – WoolierThanThou Nov 25 '19 at 9:09
  • $\begingroup$ @WoolierThanThou I think the hypothesis is that $\int fg$ exist and $|\int fg| \leq 1$ whenvever $g $ is smooth with compact support. $\endgroup$ – Kavi Rama Murthy Nov 25 '19 at 10:10
  • $\begingroup$ @WoolierThanThou Let me assume that $f$ is continuous on $\Omega$. I have changed the post. $\endgroup$ – Binjiu Nov 25 '19 at 11:03
  • $\begingroup$ what is an "smooth domain"? A convex one? $\endgroup$ – Masacroso Nov 25 '19 at 11:09
  • $\begingroup$ @Masacroso: It is a domain such that $\partial \Omega$ is a $C^{\infty}$-manifold. $\endgroup$ – WoolierThanThou Nov 25 '19 at 11:13
2
$\begingroup$

Consider that \begin{align*} T_{f}:C_{0}^{\infty}\rightarrow\mathbb{C},~~~~g\rightarrow\int fg, \end{align*} then by assumption $\|T_{f}(g)\|\leq\|g\|_{L^{2}}$ and hence $\|T_{f}\|\leq 1$. But $C_{0}^{\infty}$ is dense in $L^{2}$, so there is a unique extension $\overline{T}\in(L^{2})^{\ast}$ of $T_{f}$ such that $\left\|\overline{T}\right\|=\|T_{f}\|$. By Riesz Representation Theorem, we have $(L^{2})^{\ast}=L^{2}$ in the sense that a unique $h\in L^{2}$ is such that \begin{align*} \overline{T}(g)=\int hg,~~~~g\in L^{2}, \end{align*} and that $\|h\|_{L^{2}}=\left\|\overline{T}\right\|$. It follows that $\|h\|_{L^{2}}\leq 1$ and that \begin{align*} \int(h-f)g=0,~~~~g\in C_{0}^{\infty}. \end{align*} If it were shown to be the case that $h-f=0$ a.e. then we are done.

So the matter is now to show that $\displaystyle\int fg=0$ for all $g\in C_{0}^{\infty}$ will imply that $f=0$ a.e.

First note that the existence of $\displaystyle\int fg$ entails that $\displaystyle\int|fg|<\infty$. For a fixed compact set $K$, take a nonnegative $g\in C_{0}^{\infty}$ such that $g=1$ on $K$, then $\displaystyle\int|fg|\geq\int_{K}|f|$, then $f\in L^{1}(K)$.

On the other hand, for a fixed $x$, we have $\displaystyle\int f(\cdot)\varphi_{\epsilon}(x-\cdot)=0$, where $\varphi_{\epsilon}$ is a standard nonnegative mollifier, the equation is no more than saying that $\varphi_{\epsilon}\ast f(x)=0$. As $\varphi_{\epsilon}\ast f\rightarrow f$ in $L^{1}(K)$, we have $f=0$ a.e. on $K$.

The result follows by considering an exhaustion of compact sets to the whole space.

$\endgroup$
0
$\begingroup$

Okay, I seem to have a proof that works assuming that $f\in L^{\infty}_{loc}(\Omega)$ (hence, in particular, if $f\in C(\Omega)$).

Indeed, let $\varphi\in C^{\infty}_0(\Omega)$ and let $(\eta_{\varepsilon})_{\varepsilon\in(0,1]}$ be a (positive) mollifier. Then, since $f\in L^{\infty}(\textrm{supp}(\varphi)),$ we have

$$ \left| \int_{\Omega} f^+ \varphi \right|=\left|\int f \varphi 1_{\{f>0\}}\right|=\lim_{\varepsilon\to 0^+} \left|\int f (\varphi 1_{\{f>0\}} *\eta_{\varepsilon}) \right|\leq \lim_{\varepsilon\to 0^+}||\varphi 1_{\{f>0\}}*\eta_{\varepsilon}||_2=||\varphi 1_{\{f>0\}}||_2\leq ||\varphi||_2, $$ where we use that $\varphi 1_{\{f>0\}} *\eta_{\varepsilon}\in C_0^{\infty}(\Omega)$ for $\varepsilon$ sufficiently small (we extend by $0$ to $\mathbb{R}^n$ when defining the convolution). We get that $f^{+}$ and, similarly, $f^{-}$ must satisfy the same property.

Hence, we can assume that $f$ is positive. Let $(K_n)_{n\in \mathbb{N}}$ be an exhaustion of $\Omega$ by compacts and let $f_n=f1_{K_n}.$ Then, $f_n\in L^{\infty}(K_n)\subseteq L^{2}(K_n)$ and

$$ ||f_n||_{L^2}^2= \int f_n^2=\lim_{\varepsilon\to 0^+} \int f_n (f_n*\eta_{\varepsilon})\leq \limsup_{\varepsilon\to 0^+} \int f (f_n*\eta_{\varepsilon})\leq \lim_{\varepsilon\to 0^+} ||f_n*\eta_{\varepsilon}||_{L^2}=||f_n||_{L^2}, $$ implying that $||f_n||_{L^2}\in [0,1]$. Applying monotone convergence, we get that $f\in L^2(\Omega)$ and $||f||_{L^2}\leq 1.$

$\endgroup$
3
  • $\begingroup$ Actually, I guess I'm just using that $f\in L^2_{loc}(\Omega)$. $\endgroup$ – WoolierThanThou Nov 25 '19 at 12:00
  • $\begingroup$ For $f\in L_{\text{loc}}^{2}$ can you really do $\displaystyle\int f\varphi 1_{\{f>0\}}=\lim_{\epsilon}\displaystyle\int f(\varphi 1_{\{f>0\}}\ast\eta_{\epsilon})$? $\endgroup$ – user284331 Nov 25 '19 at 16:28
  • $\begingroup$ Yes: $|| f(\varphi 1_{\{f>0\}}- \varphi 1_{\{f>0\}}*\eta_{\varepsilon})||_{L^1}\leq ||f (1_{\varphi *\eta_{\varepsilon}\neq 0}+1_{\varphi\neq 0})||_{L^2} ||\varphi 1_{\{f>0\}}-\varphi 1_{\{f>0\}}*\eta_{\varepsilon}||_{L^2}$ by Cauchy Schwarz, and $g *\eta_{\varepsilon}\to g$ in $L^p$ for every $p$ such that $g\in L^p$ (unless of course $p=\infty$, in which case, we need $g$ continuous). $\endgroup$ – WoolierThanThou Nov 25 '19 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.