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Is a bounded continuous function defined on $\Bbb R$ differentiable? Why so?

The query is fueled by the following question:

Let $f : \Bbb R \rightarrow \Bbb R$ be a bounded continuous function. Define $ g : [0,\infty) \rightarrow \Bbb 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 $g$.

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Perhaps you should clarify your question. You're asking if all bounded continuous functions defined on $\mathbb R$ are differentiable, but so far two users have answered the question which prompted the first question mentioned on this comment. – Git Gud Apr 30 '14 at 11:26
I require a head-start to the problem I've given. In it's connection, I had the former question. But it would be nice to get some help on the latter part as well. – A.Chakraborty Apr 30 '14 at 19:38

Not in general, no. For example, the function $$ f(x) = \left\{ \begin{array}{ll} |x|, x \in [-1,1] \\ 1 \quad \text{otherwise} \end{array} \right.$$ is not differentiable for $x \in \{ -1,0,1\}.$ Your function $g$ can be rewritten as $$g(x) = 2x\int_{0}^{x} tf(t) \, dt - 2x\int_{0}^{-x} tf(t) \, dt + \int_{0}^{x} f(t) \, dt - \int_{0}^{-x} f(t) \, dt$$ Each term on the right is differentiable due to the (first) fundamental theorem of calculus, for which to apply it is necessary to assume that $f$ is bounded and continuous.

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If function $h$ is continuous on $\mathbb{R}$ then $k_{1}\left(x\right)=\int_{0}^{x}h\left(t\right)dt$ and $k_{2}\left(x\right)=\int_{-x}^{0}h\left(t\right)dt$ are both differentiable on $\left(0,\infty\right)$ with $k_{1}'\left(x\right)=h\left(x\right)$ and $k_{2}'\left(x\right)=h\left(-x\right)$.

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