Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The exercise is about convex functions:

How to prove that $f(t)=\int_0^t g(s)ds$ is convex in $(a,b)$ whenever $0\in (a,b)$ and $g$ is increasing in $[a,b]$?

I proved that

$$f(x)\leq \frac{x-a'}{b'-x}f(b')+\left(1-\frac{x-a'}{b'-x}\right)f(a')$$

when we have


share|improve this question
$b'-x$ should be $b'-a'$. –  Yimin May 11 '12 at 0:38
Yeah, but I couldn't pass that stage, I just came 'till there. –  André Lima May 11 '12 at 0:41
The converse is true also. –  copper.hat May 11 '12 at 7:02

3 Answers 3

up vote 2 down vote accepted

I also overcome with a different approach. Assume initially $g(t)\geq 0$.

We will consider the auxiliary function $\int_{a'}^x g(t)dt$ to that this function satisfies the convexity conditions in $a'<x<b$, that is,

$$\int_{a'}^x g(t)dt\leq \frac{x-a'}{b'-a'}\left(\int_{a'}^x g(t)dt+\int_x^{b'} g(t)dt\right).$$

To prove that observe

$$\int_{a'}^x g(x)dt (1-\frac{x-a'}{b'-a'})=\frac{x-a'}{b'-a'}\int_{x}^{b}g(x)dt$$

(Note that the integration is in t and g(x) is constant in that

Because $g$ is increasing we have $\int_{a'}^x g(t)dt\leq \int_{a'}^x g(x)dt$ and $\int_{x}^{b'} g(x)dt\leq \int_{x}^{b'} g(t)dt$

Then $$\int_{a'}^x g(t)dt (1-\frac{x-a'}{b'-a'})\leq\frac{x-a'}{b'-a'}\int_{x}^{b}g(t)dt$$

Rearranging it is the desired result.

share|improve this answer

You could use the fact that a function $f$ is convex on a nonempty, open interval $(a,b)$ if and only if $$ {f(x)-f(c)\over x-c}\le {f(d)-f(x)\over d-x} $$ whenever $a<c<x<d<b$. This follows from the fact that a chord through two points on the graph of a convex function lies on or above the graph of the function.

Then you need to show that, given $a<c<x<d<b$, $$\tag{1} {\int_c^x g(x)\,dx\over x-c}\le{\int_d^x g(x)\,dx\over d-x}. $$ But, since $g$ is increasing, there are numbers $e$ and $f$ with $g(c)\le e\le g(x)$ and $g(x)\le f\le g(d)$ such that $e={\int_c^x g(x)\,dx\over x-c}$ and $f={\int_d^x g(x)\,dx\over d-x}$; which shows that $(1)$ indeed holds.

share|improve this answer

A slightly different approach:

We need to show $f(x+\lambda(y-x)) \leq f(x) + \lambda (f(y)-f(x))$, with $\lambda \in (0,1)$. Suppose $x<y$. Then $$f(x+\lambda(y-x)) - f(x) = \int_{x}^{x+\lambda(y-x)} g(s) \; ds$$ Using the change of variables $t=\frac{s-x}{\lambda}+x$, we get $$\int_{x}^{x+\lambda(y-x)} g(s) \; ds = \int_{x}^{y} g(\lambda(t-x)+x) \; \lambda \; dt \leq \lambda \int_{x}^{y} g(t) \; dt = \lambda(f(y)-f(x)),$$ where the second to last step follows because $\lambda(t-x)+x \leq t$, and $g$ is increasing.

If $x>y$, let $\mu = 1-\lambda$ (note $\mu \in (0,1)$), then we have already shown that $$f(y+\mu(x-y)) \leq f(y) + \mu (f(x)-f(y)).$$ Since $c+\mu(d-c) = c+(1-\lambda)(d-c) = d+\lambda(c-d)$, the desired result follows.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.