Prove that there exists an interval $[u,v]\subset [a,b]$ where $f(x)\geq \frac{1}{b-a} \int^b_a f(y) \ \mathrm{d}y \quad \forall x \in [u,v]$.

Question

I was tryng to prove that if $f$ is continous on $[a,b]$ there exists an interval $[u,v]\subset [a,b]$ where $$f(x)\geq \frac{1}{b-a} \int^b_a f(y) \ \mathrm{d}y \quad\forall x \in [u,v].$$

I saw this answer where it is stated that

Since $\hat{f}$ is continuous at some $c \in (a,b)$, we can find $\varepsilon > 0\,$ so that $$\hat{f}(x) > \frac 1{b-a} \int_a^b \hat{f}(y)\,dy + \varepsilon$$ for all $x \in (c-\varepsilon,c+ \varepsilon) \subset (a,b)$.

In my case $f$ is continous everywhere, but it is not obvious to me why we can find such an $\varepsilon$.

Answer 1

Consider a proof by contradiction. Let, on the contrary, $$f(x)\lt\frac{1}{b-a} \int^b_a f(y) \ \mathrm{d}y \quad\forall x \in [a,b]\tag{1}$$ Suppose in $[a, b]$, the maximum value of $f(x)$ is $ M$ at $x=x_0$. Then, since $(1)$ holds $\forall x \in [a,b]$, it also holds for $x=x_0$ and thus $$M < \frac{1}{b-a} \int^b_a f(y) \ \mathrm{d}y$$$$ \leq\frac{1}{b-a}\cdot \int^b_a M \ \mathrm{d}y = \frac{1}{b-a}\cdot (b-a)M = M$$, thus giving $M<M$, which is a contradiction. $\blacksquare$