Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $g$.

share|improve this question
    
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 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 at 19:38

2 Answers 2

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.