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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be continuous and let $a$ be a nonzero real number. Show that the function

$F(x)=\frac{1}{2a}\int_{-a}^{a}{f(x+t)dt}$ is differentiable and has continuous derivative.

My thoughts were that

$\frac{d}{dx}(\frac{1}{2a}\int_{-a}^{a}{f(x+t)dt})= \frac{1}{2a}\frac{d}{dx}(\int_{-a}^{a}{f(x+t)dt}) = \frac{1}{2a}(f(x+a)-f(x-a))$, which is continuous.

Is this correct? It seemed a little too easy.

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    Well, some care is needed. Directly interchanging the differentiation and integration operators would require $f$ to be differentiable (at least). Siminore's answer avoids this.2012-08-21

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Set $x+t=u$, so that $t=a$ means $u=x+a$ and $t=-a$ means $u=x-a$. Then $ F(x)=\frac{1}{2a}\int_{x-a}^{x+a} f(u)\, du. $ You can now apply the fundamental theorem of calculus.

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    Well, you did not explain the second identity...2012-08-21
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You are given that $f(x)$ is continuous, but you are not given that $f_{x}(x,t)$ is continuous which is a requirement in Leibniz rule. The rule for interchanging differentiation with integration the way you did. So just follow Siminore's answer.

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    This might be more a comment than an answer.2012-08-22