Prove an integral inequality $|\int\limits_0^1f(x)dx|\leq\frac{1-a+b}{4}M$

Question

Let $f$ be a differentiable function on $[0,1]$ and $a,b\in(0,1)$ such that $a<b$, $\int\limits_0^af(x)dx=\int\limits_b^1f(x)dx=0$. Show that: $$\left|\int_0^{1} f(x)\,dx\,\right|\leq\frac{1-a+b}{4}\,M$$ where $M=\sup\limits_{x\in[0,1]}|f'(x)|$.

Answer 1

Since $f$ is continuous, it must have zeroes in both intervals $[0,a]$ and $[b,1]$. By the Mean value theorem, we deduce that $$ |f(a)| \leq Ma,\qquad |f(b)| \leq M(1-b). $$

By the Mean value theorem again, there exists $c \in (a,b)$ such that $$\frac{1}{b-a}\left[\int_a^bf(x)dx - \frac{1}{2}(b-a) [f(a)+f(b)]\right] = -(c-\frac{a+b}{2})f'(c). $$ Therefore, $$ \left|\int_a^b f(x)dx\right| \leq \frac{b-a}{2} \left(|f(a)| + |f(b)| + (b-a)M\right) \leq \frac{b-a}{2}M. $$ The conclusion follows since $b-a < 1$ implies $\frac{b-a}{2} < \frac{b-a + 1}{4}$ and $$ \int_0^1 f(x)dx = \int_a^b f(x)dx. $$