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Let $f$ be an integrable function on $\mathbb{R}$, and $g$ be an $L^\infty$ function on $\mathbb{R}$.

Then, the convolution $f*g$ is said to be continuous and bounded on R. I managed to show that it is bounded, but can't show that it is continuous. The convolution is defined as an integral, so it seems intuitively clear that it is continuous. Could anyone show me how to rigorously prove this?

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marked as duplicate by GEdgar, user147263, rlartiga, Davide Giraudo real-analysis Apr 27 '15 at 19:41

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Notice that for $ f \in {L^{1}}(\Bbb{R}) $ and $ g \in {L^{\infty}}(\Bbb{R}) $, we have $$ \forall x_{1},x_{2} \in \Bbb{R}: \quad (f \star g)(x_{1}) - (f \star g)(x_{2}) = \int_{\Bbb{R}} [f(x_{1} - y) - f(x_{2} - y)] g(y) ~ \mathrm{d}{y}. $$ Using this, we can show that if $ f \in {C_{c}}(\Bbb{R}) $, then the result is true by the Lebesgue Dominated Convergence Theorem. Finally, use the denseness of $ {C_{c}}(\Bbb{R}) $ in $ {L^{1}}(\Bbb{R}) $.

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