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Suppose that $f$ and $g$ are continuous in $\mathbb{R}^n$. Then $$ \int_{\mathbb{R}^n} f(x)\phi(x)dx=\int_{\mathbb{R}^n} g(x)\phi(x)dx, \forall\phi\in \mathcal{D}$$ implies that $f(x)=g(x),\forall x\in\mathbb{R}^n$.

All kinds of variants of this question have been asked on this site before, f.ex. If $f\in L^1(\mathbb{R})$ is such that $\int_{\mathbb{R}}f\phi=0$ for all continuous compactly supported $\phi$, then $f\equiv 0$. and If $f\in L^1_{loc}(\mathbb{R})$ and $\int f\varphi=0$ for all $\varphi$ continuous with compact support, then $f=0$ a.e. . But what about simply continuous functions $f$ and $g$? Every continuous function is locally integrable, so by the proof provided in the second link (generalized to $\mathbb{R}^n$), we would have that $f-g=0$ a.e. in $L^1$. How do I actually show that $f=g$ everywhere in $\mathbb{R}^n$?

Is there another way to prove this, without using local integrability?

Thanks.

MyWorld
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2 Answers2

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Since $f$ and $g$ are continuous, $f-g$ is also continuous. By linearity, this means that you should verify whether $\int_{\mathbb{R}^n} f(x)\phi(x)dx=0$ for all continuous compactly supported functions implies $f\equiv0$.

Assume $f(x_0)> 0$, then by continuity there exists a ball $B_\delta(x_0)$ such that $f(x)>\frac{1}{2}f(x_0)$ for all $x$ in said ball. There is also a ball $B_{\delta'}(x_0)\supsetneq B_{\delta}(x_0)$ for which $f(x)>0$. Take a continuous function $\phi$ such that $\phi_{B_\delta(x_0)}=1$ and $\phi_{\big( B_{\delta'}(x_0) \big)^c}=0$. Check the integral $\int_{\mathbb{R}^n}f(x)\phi(x)dx$.

Keen-ameteur
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  • Very elegant solution! As for the integral check: $\int_{\mathbb{R}^n} f(x)\phi(x)dx=\int_{B_{\delta'}(x_0)} f(x)\phi(x)dx > \int_{B_{\delta}(x_0)} f(x)dx > \frac12 f(x_0) m(B_{\delta}(x_0)) > 0$, which contradicts the assumption, so $f=0$ everywhere on $\mathbb{R}^n$. Thank you! – MyWorld Nov 18 '20 at 13:11
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The complement of a measure zero set in $\mathbb{R}$ is of course dense. So if $f-g$ is continuous and equal to $0$ on a dense set, it must be zero everywhere.