Consider a measurable function $f:[0,1]\rightarrow \mathbb{R}$ such that $f(x)\geq ||f||_2$. So I am stuck on this question and I don't really know how to go about it. The question is 
Let $X=[0,1]$ and $\mu = \lambda$ be the Lebesgue measure. 
Consider a measurable function $f:[0,1]\rightarrow \mathbb{R}$ such that $f(x)\geq ||f||_2$ for all $x \in [0,1]$. Prove that there exists an $x \in [0,1]$ such that $f(x_0) =||f||_2$.
So far, I have written out the statements
$||f||_2=\bigg(|f|^2 d\lambda \bigg)$ and $f(x)\geq \bigg( \int |f|^2 d\lambda \bigg)^{\frac{1}{2}}$

Using Holder's inequality
$$
\|f\|_2 = \int_0^1 \|f\|_2 \leq \int_0^1 f \leq \left(\int_0^1 |f|^2\right)^{1/2} = \|f\|_2.
$$
Hence the non-negative function $g(x) := f(x) - \|f\|_2$ satisfies $\int_0^1 g = 0$, so that $g(x) = 0$ a.e. in $[0,1]$, i.e. $f(x) = \|f\|_2$ a.e. in $[0,1]$.
