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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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You are missing some steps here. You haven't shown why the last equality is true. See Siminore's answer below. – copper.hat Aug 21 '12 at 15:12
Yes, this is correct. You might want to explain why it is that you can interchange the integration and differentiation operations. – Sasha Aug 21 '12 at 15:12
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. – copper.hat Aug 21 '12 at 15:15
up vote 3 down vote accepted

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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So is my answer correct? Because I think I did what you just said to get from the middle step to the last step. – neelp Aug 21 '12 at 15:11
Well, you did not explain the second identity... – Siminore Aug 21 '12 at 15:12

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. – Did Aug 22 '12 at 8:31

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