# convexity problem

• May 22nd 2008, 12:22 AM
szpengchao
convexity problem
• May 22nd 2008, 12:44 AM
Isomorphism
Quote:
For the first part assuming $0 < t$,

$f''(t) \geq 0 \Rightarrow \int_0^t f''(t)\, dt \geq 0 \Rightarrow f'(t) \geq f'(0) > 0 \Rightarrow f'(t) > 0$

Rinse and repeat:

$f'(t) > 0 \Rightarrow \int_0^t f'(t)\, dt > 0 \Rightarrow f(t) > f(0)= 0 \Rightarrow f(t) > 0$

Now for $t_2 > t_1, f'(t) > 0 \Rightarrow \int_{t_1}^{t_2} f'(t)\, dt > 0 \Rightarrow f(t_2) > f(t_1)$

So f is monotonically increasing and yet f(0) = f(1) = 0. Thus f is identically 0 throughout [0,1]. Similarly for the next one, except

$\int_{t}^1 f''(t) \geq 0 \Rightarrow f'(t) \leq f'(1) \leq 0 \Rightarrow f'(t) < 0$

Then you can repeat what you did before :)
• May 22nd 2008, 12:49 AM
Moo
Hello,

An insignificant report...

Quote:

$
f''(t) \geq 0 \Rightarrow \int_0^t f''(t)\, dt \dots
$

I'm not sure it's correct to call the variable t in the integrand, just like the variable in the boundaries... I know it's understandable, but well, it's quite disturbing
• May 22nd 2008, 12:53 AM
Isomorphism
Quote:

Originally Posted by Moo
Hello,

An insignificant report...

I'm not sure it's correct to call the variable t in the integrand, just like the variable in the boundaries... I know it's understandable, but well, it's quite disturbing

I dont understand whats the problem..... Do you mean you dont want the same variable appearing in both places?

Here its not troublesome, so why bother inventing a new one?
• May 22nd 2008, 12:56 AM
Moo
Quote:

Originally Posted by Isomorphism
I dont understand whats the problem..... Do you mean you dont want the same variable appearing in both places?

Here its not troublesome, so why bother inventing a new one?

Nevermind, forget it (Wink)
• May 23rd 2008, 05:19 AM
szpengchao
any other way
any other way of doing this by convex function's property?

and i hav a question, is there any property of convexity in R^n ?

f''(x)>=0 gives global definition in R
but this fails in R^n, why???