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

Question

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 $g$.

Answer 1

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.

Answer 2

Hint:

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 $k_{2}'\left(x\right)=h\left(-x\right)$.