# A question on definite integral

Suppose $f$ is increasing on $[0,\infty)$ and $\forall x: f(x)>0 \ g(x)=\frac{1}{x}\int_{0}^{x}f(u)du$ for $0<x<\infty$ then which of the following is true assuming $\forall x\in (0,\infty)$?

1. $g(x)\le f(x)$
2. $xg(x)\le f(x)$
3. $xg(x)\ge f(0)$
4. $yg(y)-xg(x)\le (y-x)f(y)$, $x<y$
-

Another way to look at it:
$xg(x)$ is the total area under the curve $f(x)$ within the interval $(0,x)$. So $g(x)$ is the average height of $f(x)$ within the interval $(0,x)$. When you see $xg(x)$, think of it as an area; when you see $g(x)$, you're dealing with an average height.

So, follow your intuitive, devise various graphs or functions that violates the options.

For option 4, transform it, rearrange the terms so that it looks like something prettier and makes sense.

-
@Mex A line of apology for the homework assertion. But pointing out where you'd like to know gives answerers a clear direction. :) – FrenzY DT. Jun 3 '12 at 13:38

It looks like homework, so I won't do it completely.

1. Consider the upper bound for $\int_0^x f(u)du$. Since $f$ is increasing, this is trivial.
2. Try $f(x) = x + 1$
3. Consider what happens when $x$ is small. Notice that $f(0) = \int_0^1 f(0)du$.
4. If you write $y - x = \epsilon$, then the inequality is $\int_x^{x+\epsilon}f(u)du \leq \epsilon f(y)$. Consider the upper bound for the integral as you did in part 1.
-
I was almost going to submit a full answer, glad that you posted (+1) just before I did :-). – TenaliRaman Jun 3 '12 at 13:05
@Karolis, tenali, this is not a home work as your babies are doing. – La Belle Noiseuse Jun 3 '12 at 13:26