# Convolution of an integrable function an $L^\infty$ function [duplicate]

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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?

## marked as duplicate by GEdgar, user147263, rlartiga, Davide Giraudo real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 27 '15 at 19:41

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## 1 Answer

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 ${L^{1}}(\Bbb{R})$.