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- May 21st 2008, 11:22 PMszpengchaoconvexity problem
- May 21st 2008, 11:44 PMIsomorphism
For the first part assuming $\displaystyle 0 < t$,

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

Rinse and repeat:

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

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

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

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

Then you can repeat what you did before :) - May 21st 2008, 11:49 PMMoo
Hello,

An insignificant report...

Quote:

$\displaystyle

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

$

- May 21st 2008, 11:53 PMIsomorphism
- May 21st 2008, 11:56 PMMoo
- May 23rd 2008, 04:19 AMszpengchaoany 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???