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Show that if $f \in L^{\infty}$ and $g \in L^1$ ,then $f \star g$ is continuous.

Ideas: assume $g$ is continuous and compactly supported, if not we can just approximate by a sequence of compactly supported continuous functions. Let $(x_n) \rightarrow x$ and consider $(f \star g)(x_n)$. Let $h_n(y) = g(x_n-y)$. We have that $h_n(y) \rightarrow g(x-y)$ pointwise. And finally monotone converge theorem gives that $(f \star g)(x_n) \rightarrow (f \star g)(x)$. Is this on the correct path?

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  • $\begingroup$ I don't see why invoking the monotone convergence theorem is valid here: What sequence are you applying it to? $\endgroup$ – user61527 Mar 23 '14 at 19:40
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Let $\epsilon > 0$ and let $h(x)$ be a continuous function such that $||h-g||_1 < \epsilon$ with compact support (such functions are dense in $L^1$).

Now, $|(f \star g) (x_1) - (f \star g) (x_2)| \leq \int |f(y)||g(x_1 -y) - g(x_2 - y)| dy$

$\leq \int |f(y)||g(x_1 -y) - h(x_1 - y)|dy $ $+ \int |f(y)||h(x_1 -y) - h(x_2 - y)|dy $ $ + \int |f(y)||h(x_2 -y) - g(x_2 - y)|dy$

$\leq ||f||_\infty 2 \epsilon + ||f||_\infty\int|h(x_1 -y) - h(x_2 - y)|dy$ but $h$ is continuous with compact support, so there exists a $\delta > 0$ such that $\int |h(x_1 -y) - h(x_2 - y)|dy < \epsilon$ when $|x_1 - x_2| < \delta$.

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