1. Dec 26, 2012

### Felafel

1. The problem statement, all variables and given/known data
Let $f: [0, a]$ ---> $\mathbb{R}$ be positive and increasing.
Prove that the function G, such that:
$G(x):= \frac{1}{x} \int_0^x f(t)dt$ $x\in (0,a)$
is increasing.

3. The attempt at a solution
I know that if the first derivative of a function is positive, that function is increasing.
By definition of primitive function, we have that the derivative of $\int_0^x f(t)dt$ is $f$, which is positive by hypothesis.
However, i can't handle that $\frac{1}{x}$. If I derivate it I get $\frac{1}{x^2}$, making things worse.
What should I do with it?

2. Dec 26, 2012

### awkward

Have you tried to write out a formula for $G'(x)$? (Apply the rule for the derivative of a product.)

3. Dec 26, 2012

### Felafel

yes, and i get $G ' (x)= -\frac{1}{x^2} \int_0^x f(t)dt + \frac{1}{x} f(x)+c$
$= \frac{1}{x}(-\frac{1}{x}\int_0^x f(t)dt+f(x)+c)$
but there appears a minus, and i don't know how to go on with the proof

4. Dec 26, 2012

### Felafel

I've had an idea, maybe it's correct:
Let us consider $G ' (x)$
$\frac{1}{x}$ is a positive number (because x>0 in the hypothesis), so i can do as if it weren't there.
Now,
$-\frac{1}{x} \int_0^x f(t)dt+f(x)+c > 0$ if $f(x)+c > \frac{1}{x} \int_0^x f(t)dt$
But $f(x)$ is the upper bound, so for the integral mean theorem we have:
$f(x) (\frac{1}{x-0}) > \int_0^x f(t)dt> m (\frac{1}{x-0})$
and therefore:
$f(x)+c> (\frac{1}{x})\int_0^x f(t)dt$
QED.
or have I made any mistakes?

5. Dec 26, 2012

### Fightfish

For simplicity, let the antiderivative of f be F.
Then,
$$-\frac{1}{x}\int_0^x f(t)dt+f(x) = -\frac{1}{x} (F(x) - F(0)) + f(x)$$
We can apply the mean value theorem in reverse:
There exists a point in the interval [0,x] (lets call it c) such that $$F'(c) = f(c) = \frac{F(x) - F(0)}{x-0}$$
Then,
$$-\frac{1}{x} (F(x) - F(0)) + f(x) = -f(c) + f(x)$$
But f is an increasing function, so f(x) > f(c) because c < x.

6. Dec 26, 2012

### Felafel

Great!
Do you think my version above could work as well?