Proving Inequality for Continuously Differentiable Functions on Closed Interval

[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.

 

[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)

 

[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.

 

[dirk_mec1]

Look at this:

<br /> |f(t)| = \left|\int_0^t f&#039;(x) \mbox{ d}x \right| \leq \int_0^t |f&#039;(x)|\ \mbox{ d}x \leq \int_0^1 |f&#039;(x)|\ \mbox{ d}x <br />

Can you finish this?