Solving for c: Is it Always Possible?

[dirk_mec1]

Homework Statement

http://img143.imageshack.us/img143/1861/14726023et5.png

Can I take c such that

<br /> c = \frac{2 ||f ||_{\infty} }{ ||f||_E}<br />

and in case f= 0 everywhere c=1?

 

[Dick]

The point to c being a constant is that it's independent of f, right? Try and read the question eliminating the 'norm' words. You have a function f(x) that has a bounded derivative on [0,1] and f(0)=0. Can you get a bound for f(x) in terms of the bound on the derivative?

 

[dirk_mec1]

The point to c being a constant is that it's independent of f, right?

Yes, you're right!

Try and read the question eliminating the 'norm' words. You have a function f(x) that has a bounded derivative on [0,1] and f(0)=0. Can you get a bound for f(x) in terms of the bound on the derivative?

Well I know that:

f&#039;(x)= \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}

But I don't see how that is going to help...

 

[Dick]

If you are studying Banach spaces, you must know more about derivatives than just the definition. If you know f'(x) how can you find f(x)?

 

[dirk_mec1]

If you are studying Banach spaces, you must know more about derivatives than just the definition. If you know f'(x) how can you find f(x)?

f(x) = \int_0^x f&#039;(t)\ \mbox{d}t

I carefully looked two times in my notes from my instructor and I can't find anything that relates f to f' (in context of Banach spaces). Can you give please me another hint, Dick?

 

[morphism]

f(x) = \int_0^x f&#039;(t)\ \mbox{d}t

Take the absolute value of both sides of this, and then work from there.

 

[dirk_mec1]

Do you mean like this?

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

Is this correct?

 

[Dick]

Do you mean like this?

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

Is this correct?

That's what I've been waiting for.