True of false? If $f:[a,b]\to \mathbb{R}$ is decreasing, then $g:(a,b)\to \mathbb{R}:~g(x)=\frac{1}{x-a}\int\limits_a^xf(t)\,\mathrm{d}t$ is decreasing.

Attempt. Since $f$ is monotonic, $f$ is integrable and $g$ is well defined. If $f$ was continuous, then $g$ would be differentiable and: $$g'(x)=\frac{f(x)(x-a)-\int\limits_a^xf(t)\,\mathrm{d}t}{(x-a)^2}= \frac{f(x)(x-a)-f(\xi)\,(x-a)}{(x-a)^2}=\frac{f(x)-f(\xi)}{x-a}\leqslant 0$$ for some $\xi\in (a,x)$ by MVT for integrals (and so $f(x)\leqslant f(\xi)$ for the decreasing $f$).

But happens when $f$ is not assumed continuous?

Thanks in advance.

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    $\begingroup$ Fixed $g'(x)$ in your equation. But did you also mean $\dfrac{f(x)-f(\xi)}{x-a}\leq0$ instead? $\endgroup$ Oct 25, 2019 at 17:52
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    $\begingroup$ Think discretely. If you were to introduce an element into a set that is lower than the average value of the original set, would that increase or decrease the average? $\endgroup$ Oct 25, 2019 at 17:59
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    $\begingroup$ $f(x)(x-a)$ Is the integral of the constant value $f(x)$ on the interval $(a,x).$ $\endgroup$
    – William M.
    Oct 25, 2019 at 19:37

1 Answer 1


$g(x)$ is the average of $f$ on the interval $[a,x]$. If $a < x < y < b$, $$ \frac{1}{x-a} \int_a^x f(t)\; dt \ge f(x) \ge \frac{1}{y-x} \int_x^y f(t)\; dt$$ so $$\eqalign{\frac{1}{y-a}\int_a^y f(t)\; dt &= \frac{1}{y-a} \int_a^x f(t)\; dt + \frac{1}{y-a}\int_x^y f(t)\; dt\cr &\le \frac{1}{y-a} \int_a^x f(t)\; dt + \frac{y-x}{y-a} f(x)\cr &\le \frac{1}{y-a}\int_a^x f(t)\; dt + \frac{y-x}{(y-a)(x-a)} \int_a^x f(t)\; dt\cr &= \frac{1}{x-a} \int_a^x f(t)\; dt}$$

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    $\begingroup$ The first inequality is false. In fact, it holds that $\frac 1{x-a}\int_a^xf(t)\,dt\ge f(x)$. And that's exactly what you use later on. $\endgroup$
    – amsmath
    Oct 25, 2019 at 18:11
  • $\begingroup$ Thanks, editing. $\endgroup$ Oct 25, 2019 at 19:34

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