Undergrad Proving that convexity implies second order derivative being positive

[hmparticle9]

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt:

Convexity says that
$$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$
$$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$

We know from the intermediate value theorem that there exists a $c \in (b,a)$ such that
$$\frac{f(a) - f(b)}{a-b} = f'(c).$$

Hence
$$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$
$$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)} \leq f'(c))$$
taking the limit $\lambda \rightarrow 0$
$$f'(b) \leq f'(c)$$

This is not enough. I need to show that the above holds for all $c$. I have shown that there exists one such c. Could I have some hints please? What I am getting at is that a function $f$ such that $f''(x)>0$ for all $x$ has the property that $f'(x) > f'(y)$ if $x>y$

 

[div_grad]

taking the limit $\lambda \rightarrow 0$

Why do you take the limit $\lambda \rightarrow 0$, instead of letting the difference $h \equiv b-a$ go to zero?

 

[hmparticle9]

What difference would that make?

 

[hmparticle9]

I just realized that my proof is even more wrong. I need to show that $f'(x) > f'(y)$ for $x > y$. Sorry guys. I am well off.

 

[hmparticle9]

I think I understand where you are coming from now. If we let $a$ tend to $b$ then since $c$ is in $(b,a)$ we must have $f'(b) < f'(c)$. But is that enough?

 

[hmparticle9]

We can say
$$f'(b) < f'(b+\epsilon) < f'(b+2\epsilon) < ... < f'(c)$$
for some $b < c$.