I have recently been trying some questions on Riemann integral. Got stuck in one of the problems which says:
Suppose $f$ is an increasing real-valued function on [$0$,$\infty$] with $f$($x$) $\gt$ $0$ for all $x$ and let $g$($x$)=$\frac{1}{x}$$\int_{0}^{x}f(u)du$ where $0$ $\lt x\lt$ $\infty$.
Then $g$($x$) $\le$ $f$($x$) for all $x$ in ($0,\infty$).
How to show this?
I can show that $g$($x$) $\gt$ $0$ but unable to show the above relation.
Help, please!
Let $ x>0$.
$ f $ is increasing at $ [0,x] \;\;\implies$
$$(\forall u\in[0,x]) \;\; f(u)\le f(x) \;\;\implies $$
$$\int_0^xf(u)du\le \int_0^xf(x)du \;\; \implies$$
$$\int_0^xf(u)du\le f(x)(x-0)\;\; \implies$$
$$g(x)\le f(x)$$
We used the fact that a monotonic function at $ [a,b] $ is Riemann-integrable at $[a,b]$.