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If $\phi:\mathbb R\rightarrow\mathbb R$ with, $\int_{\mathbb R}\phi(x)\ dx=1$ and $\phi_{\delta}$ is defined by, $\phi_{\delta}(x)=\frac{\phi(\frac{x}{\delta})}{\delta}$. Prove that, for every continuous function $f:\mathbb R\rightarrow\mathbb R$ with compact support,$$f*\phi_{\delta}(x)\overset{(\delta\rightarrow 0)}\longrightarrow f(x)\ ,\forall x\in\mathbb R$$ (where * is the convolution). The rate of convergence may depend on $f$ and $x$.

$\phi_{\delta}$ looks like an approximation of unity, but it is not mentioned that it has a compact support.

Since $f$ has a compact support, $\exists k>0$, s.t. $f(x)=0$, if $\lvert x\rvert>k$. Then,

$f*\phi_{\delta}(x)=\int_{\lvert y\rvert\le k}\phi_{\delta}(x-y)f(y)dy+\int_{\lvert y\rvert>k}\phi_{\delta}(x-y)f(y)dy$

the second summand is zero, so consider the first;

$\int_{\lvert y\rvert\le k}\phi_{\delta}(x-y)f(y)dy\overset{\frac{x-y}{\delta}=u}=\frac{1}{\delta}\delta\int_{u\ge \frac{x-k}{\delta}}\phi(u)f(x-\delta u)du$

$\textbf{-can somebody correct the set of the integral, over which we integrate}$ because $u\ge \frac{x-k}{\delta}≠\mathbb R$ and i guess it must be the whole space.

Can we better choose here if $\lvert y\rvert\le k$ then $x-y\le x+k$ and $u=\frac{x-y}{\delta}\le \frac{x+k}{\delta}$ which implies that $(u\le \frac{x+k}{\delta})\rightarrow\mathbb R$ as $\delta\rightarrow 0$. Can you confirm this ?

$\textbf{-but where do we use the continuity}$ and does $f(x-\delta u)$ make sense ? (already answered)

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Since $f$ is continuous you know that $f(x-\delta u)\to f(x)$ as $\delta\to 0$ for all $u\in \mathbb R$. Moreover $f$ is bounded hence $\phi(u)f(x-\delta u)$ can be dominated by $M\phi(u)$. So by Lebesgue theorem you conclude that $$ \int \phi(u) f(x-\delta u)\, du \to \int \phi(u) f(x)\, du = f(x). $$

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