# Existence of a point such that $f(c) = \sqrt{\frac{1}{b-a} \int_{a}^{b}f^{2}(x)dx}$

This has been stumping my calculus class:

If $f$ is continuous on $[a,b]$ and $f(x)\geq 0$ on $[a,b]$ show that there exists a $c$ in $[a,b]$ such that $f(c) = \sqrt{\frac{1}{b-a} \int_{a}^{b}f^{2}(x)dx}$

Thanks!

Attempts: We've tried an argument showing that you can approximate it, and even tried thinking up a function for applying the MVT or Darboux's theorem without much luck (even though we think this is the right idea)

• "If $f$ is continuous on $[a,b]$ and $f(x) \geq 0$ on $[a,b]$ show that there exists a $c$ in $[a,b]$ such that $f(c) = \sqrt{ \frac{1}{b-a} \int_a^b f^2(x) dx}$." Did I get that right? Mar 12, 2013 at 3:00
• Perfect. thanks. Mar 12, 2013 at 3:09
• Slightly more generally, with the same proof: for any continuous $f$, there exists $c$ such that $|f(c)|=\ldots$. The square root is only here to distract you. Mar 12, 2013 at 3:34

If $f$ is continuous, so is $f^2$. Let $g=f^2$. You want

$$g(c)=\frac{1}{b-a}\int_a^b g$$

Since $g$ is continuous on $[a,b]$ it attains it maximum and minimum values, $m,M$. Then $$m\leq g \leq M$$

which means $$m(b-a)\leq \int_a^b g\leq M(b-a)$$

or

$$m \leq \frac{1}{b-a}\int_a^b g \leq M$$

Since $g([a,b])=[m,M]$, $$g(c)=\frac{1}{b-a}\int_a^b g$$ for some $c\in [a,b]$.

Hint: Use the so-called "first mean value theorem for integrals" with $G(t)=f^2(t)$ (with notation as in the wikipedia entry).