# Continuously differentiable

1. Jan 3, 2009

### johnson12

I'm having trouble with this inequality:

let f be (real valued) continuously differentiable on [0,1] with f(0)=0, prove that

sup$$_{x\in[0,1]}$$ $$\left|f(x)\right|$$ $$\leq$$ $$\int^{1}_{0}\left|f\acute{}(x)\right| dx$$

Thanks for any help.

2. Jan 3, 2009

### jgens

Clarifying questions: Do you intend to find the supremum of the function or the interval? Furthermore, is the second f(x) term an f(x) or f'(x)? (It looks like there is a little tick next to it but I cannot tell for sure)

3. Jan 3, 2009

### johnson12

Sorry for the discrepancy, the problem is to show that the supremum of |f(x)| over [0,1] is less than or equal to the integral from 0 to 1 of |f ' (x)|, where f ' (x) is the derivative of f.

4. Jan 4, 2009

### dirk_mec1

Look at this:

$$|f(t)| = \left|\int_0^t f'(x) \mbox{ d}x \right| \leq \int_0^t |f'(x)|\ \mbox{ d}x \leq \int_0^1 |f'(x)|\ \mbox{ d}x$$

Can you finish this?