Integral mean of non decreasing function is non decreasing

Question

Given a continuous non decreasing function $r(t)$ prove that $\frac{1}{t}\int_{0}^{t} r(s)ds$ is also non decreasing.

Attempt

Non decreasingness for continuous functions would imply that the derivative is non negative therefore we start with $r'(t) \ge 0$ and take the derivative of the integral mean we get $r(t) - \frac{1}{t^2}\int_{0}^{t}r(s)ds$ but it's unclear to me how to prove that it is greater or equal than 0.

The claim is wrong. Take $r(t)=e^t$. You will see that the integral mean of that is $e^t/t$ which is decreasing on $t\in[0,1]$. — Kurt G., Oct 08 '22 at 12:08
Does this answer your question? If $f'$ is increasing and $f(0)=0$, then $f(x)/x$ is increasing — zwim, Oct 08 '22 at 12:13
@zwim I don't think it does, not sure if they can be connected somehow — Mattiatore, Oct 08 '22 at 12:34

score 2 · Accepted Answer · answered Oct 08 '22 at 14:20

The derivative of the integral mean is $$\tag{1} \frac{r(t)}{t}-\frac{\int_0^tr(s)\,ds}{t^2} $$ (in your corresponding formula is an error). Since $r$ is non decreasing we have $$ \frac{1}{t}\int_0^tr(s)\,ds\le\frac{1}{t}\int_0^t r(t)\,ds=r(t)\,. $$ Therefore (1) is bounded from below by $$ \frac{r(t)}{t}-\frac{r(t)}{t}=0 $$ which implies that the integral mean is non decreasing.

Integral mean of non decreasing function is non decreasing

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