Does there exist $\xi\in[a,b]$ such that $|f(\xi)|\leq\frac{1}{b-a}\int_{a}^{b}|f(x)| \ \text{d}x$?

Question

Suppose that $f$ is Riemann integrable. Does there exist $\xi\in[a,b]$ such that $$|f(\xi)|\leq\frac{1}{b-a}\int_{a}^{b}|f(x)| \ \text{d}x?$$ If there exists a number $\xi$ such that $|f(\xi)|\leq|f(x)|$ for all $x\in[a,b]$, then I think that I know how to prove it: Since $f$ is Riemann integrable, $f$ is also Lebesgue integrable. So because $|f(\xi)|1_{[a,b]}$ is a simple function bounded by $f$, it follows by definition of the Lebesgue integral of a non-negative function that $$|f(\xi)|(b-a)=|f(\xi)|\lambda([a,b])\leq\int_{[a,b]}|f| \ \text{d}\lambda=\int_{a}^{b}|f(x)| \ \text{d}x.$$ But how do I prove the general case?

Also, is there a way to avoid lebesgue integrals?

EDIT: Since $f$ is not necessarily continuous, I’m not sure if the MVT can be applied.

Answer 1

Since $f$ is Riemann integrable, $g=\lvert f \rvert$ is too. Suppose there is no such $\xi$. Then for every $y \in [a,b]$, we have $$ g(y) > \frac{1}{b-a} \int_a^b g . $$ Integrating over $[a,b]$, dividing by $b-a$, and using positivity of the Riemann integral (that is, that it preserves strict inequalities that hold everywhere) then gives $$ \frac{1}{b-a} \int_a^b g > \frac{1}{b-a} \int_a^b g , $$ a contradiction.

Answer 2

Applying the Lagrange mean value theorem to $$ F(x) = \int_a^x {f(t)dt} $$ shows that there is a $\xi \in (a,b)$ such that $$ f(\xi ) =F'(\xi)=\frac{F(b)-F(a)}{b-a}= \frac{1}{{b - a}}\int_a^b {f(t)dt} . $$ Now $$\left| {f(\xi )} \right| = \frac{1}{{b - a}}\left| {\int_a^b {f(t)dt} } \right| \le \frac{1}{{b - a}}\int_a^b {\left| {f(t)} \right|dt} . $$