# $| \int fg|\leq 1$ for any $g\in C_{0}^{\infty}$, $\|g\|_{L^{2}}=1$ then $\|f\|_{L^{2}}\leq1$

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.

• 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. – WoolierThanThou Nov 25 '19 at 9:09
• @WoolierThanThou I think the hypothesis is that $\int fg$ exist and $|\int fg| \leq 1$ whenvever $g$ is smooth with compact support. – Kavi Rama Murthy Nov 25 '19 at 10:10
• @WoolierThanThou Let me assume that $f$ is continuous on $\Omega$. I have changed the post. – Binjiu Nov 25 '19 at 11:03
• what is an "smooth domain"? A convex one? – Masacroso Nov 25 '19 at 11:09
• @Masacroso: It is a domain such that $\partial \Omega$ is a $C^{\infty}$-manifold. – WoolierThanThou Nov 25 '19 at 11:13

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.

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 $$||f||_{L^2}\leq 1.$$

• Actually, I guess I'm just using that $f\in L^2_{loc}(\Omega)$. – WoolierThanThou Nov 25 '19 at 12:00
• For $f\in L_{\text{loc}}^{2}$ can you really do $\displaystyle\int f\varphi 1_{\{f>0\}}=\lim_{\epsilon}\displaystyle\int f(\varphi 1_{\{f>0\}}\ast\eta_{\epsilon})$? – user284331 Nov 25 '19 at 16:28
• 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). – WoolierThanThou Nov 25 '19 at 16:49