A question on definite integral

Question

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. $yg(y)-xg(x)\le (y-x)f(y)$, $x<y$

Answer 1

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.

Answer 2

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.