Let $f(t):[0,t]\rightarrow \mathbb{R}$, $s\leq t$, is the following inequality true?

Question

I need to know if the following inequality is true:

Let $f(t):[0,t]\rightarrow \mathbb{R}$, $s\leq t$, then

$$\left|\max\limits_{0\leq v\leq t}\left(\int_{0}^{v}f(u)du\right)-\max\limits_{0\leq w\leq s}\left(\int_{0}^{w}f(u)du\right)\right|\leq \max\limits_{s\leq v\leq t}\left|\int_{s}^{v}f(u)du\right|$$

Answer 1

Yes it is true.

First note that using maximum instead of supremum here makes sense, because the map $v\mapsto \int_0^v f(u)du$ is continuous on a compact set $[0,t]$, and the maximum and the minimum are both attained.

Now consider two cases:
1).If $$\exists\DeclareMathOperator*{\argmax}{arg\, max}\argmax\limits_{0\leq v\leq t}\int_0^vf(u)du \in [0,s],$$ then we must have $$\exists\argmax\limits_{0\leq v\leq t}\int_0^vf(u)du=\exists\argmax\limits_{0\leq s\leq t}\int_0^vf(u)du.$$ Which makes LHS zero;
2). If $$\forall \argmax\limits_{0\leq v\leq t}\int_0^vf(u)du\in (s,t],$$ Let $v_0\in(s,t]$ be sucn an $\argmax$, and let $w_0$ be some $\argmax\limits_{0\leq w\leq s}\int_0^vf(u)du$. Then LHS will be $$|\int_{w_0}^{v_0}f(u)du|=\int_{w_0}^{v_0}f(u)du = (\int_0^s - \int_0^{w_0})f(u)du + \int_s^{v_0}f(u)ds$$ in which $(\int_0^s - \int_0^{w_0})f(u)du\le0$ and $0\le\int_s^{v_0}f(u)ds=|\int_s^{v_0}f(u)ds|\le\max_{s\leq v\leq t}|\int_{s}^{v}f(u)du|$.