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Let $f:\mathbb{R}\to \mathbb{R}$ be a bounded continuous function. Define $g:[0,\infty)\to \mathbb{R}$ by $$g(x)=\int_{-x}^{x}(2xt+1)f(t) ~dt.$$ Show that $g$ is differentiable on $(0,\infty)$ and find the derivative of $g$.

I can find the derivative. I can check differentiablity by equating left hand limit and right hand limit in one point. How can I check differentiablity on an interval? Please help.

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You need to show that $g$ is differentiable at each $x \in (0,\infty)$. –  copper.hat Aug 5 '12 at 4:30

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If $h$ is continuous on $\mathbb{R}$, then the function $\phi(x) = \int_0^x h(t) dt$ is differentiable on $\mathbb{R}$, with derivative $\phi'(x) = h(x)$. To see this, choose $ x \in \mathbb{R}$, and let $\epsilon>0$. Since $h$ is continuous, there exists a $\delta>0$ such that if $|y-x| < \delta$, then $|h(y)-h(x)| < \epsilon$. Then you have $$|\phi(x+\eta) - \phi(x) - h(x) \eta| = |\int_{x}^{x+\eta} (h(t)-h(x)) dt| \leq \epsilon \eta,$$ from which the desired result follows. Since $x \in \mathbb{R}$ was arbitrary, it follows that $h$ is differentiable on $\mathbb{R}$.

Now consider the function $\eta(x) = \int_{-x}^x h(t) dt$. Since we can write $\eta(x) = \int_0^x h(t) dt - \int_0^{-x} h(t) dt$, it is clear that $\eta$ is differentiable, and $\eta'(x) = h(x)+h(-x)$ (the latter term involves the composition rule).

Now let $x \in (0,\infty)$ and write $$g(x) = 2 x \int_{-x}^x t f(t) dt + \int_{-x}^x f(t) dt.$$

Since the functions $f$ and $t \mapsto t f(t)$ are continuous, it follows that $g$ is differentiable at $x$. Since $x$ was arbitrary, $g$ is differentiable everywhere on $(0, \infty)$. The derivative is straightforward to compute using the product rule as: $$g'(x) = 2x(x f(x)+(-x)f(-x))+2 \int_{-x}^x f(t) dt +f(x)+f(-x).$$

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