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Let $f, \phi:\mathbb{R} \rightarrow \mathbb{R}$ let $f \in C^k$, $\phi \in C_0^\infty$ and $\phi \geq 0$, $\int \phi(x)dx=1$.

Why a function $g(x,t)=\int f(x-ty)\phi(y)dy$ for $t > 0$, $x \in \mathbb{R}$, is of class $C^\infty$.

It probably follows from the following fact, however I don't know in such a way, that a convolution $(u *\phi)(x)=\int u(x-y)\phi(y)dy$, $x \in \mathbb{R}$, of every locally integrated function $u: \mathbb{R} \rightarrow \mathbb{R}$ (i.e integrated on every compact) with function $\phi \in C_0^\infty$, is a smooth function and $(u *\phi)^{(n)}=u*\phi^{(n)}$ for every $n \in \mathbb{N}$.


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up vote 3 down vote accepted

In the integral, put $u=x-ty$ (then $du=-tdy$). We get $$g(x,t)=\frac 1t\int_{—\infty}^{+\infty}f(u)\phi\left(\frac{x-u}t\right)du.$$ Put $h(x,t):=\int_{—\infty}^{+\infty}f(u)\phi\left(\frac{x-u}t\right)du$ for $x\in\mathbb R$ and $t>0$. We have to show that $h$ is $C^{\infty}$. But it's a consequence of the dominated convergence theorem, since $\phi$ has a compact support (and then all its derivatives) and $f$ is continuous on this compact (hence bounded). If we compute the derivatives on $x$ and $t$ of $\phi\left(\frac{x-u}t\right)$, we will get a function which is the integral of a continuous function with compact support.

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