# True of false? If $f$ is decreasing, then $\frac{1}{x-a}\int\limits_a^xf(t)\mathrm{d}t$ is decreasing.

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?

• Fixed $g'(x)$ in your equation. But did you also mean $\dfrac{f(x)-f(\xi)}{x-a}\leq0$ instead? Oct 25, 2019 at 17:52
• $f(x)(x-a)$ Is the integral of the constant value $f(x)$ on the interval $(a,x).$ Oct 25, 2019 at 19:37
$$g(x)$$ 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}
• 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. Oct 25, 2019 at 18:11