Weierstrass Approximation - Proof that Integral Exists

Question

Trying to answer the following question;

Let $f(x)$ be a continuous real valued function on $[0,4]$. Given any $\epsilon>0$ prove there is a polynomial $p(x)$ such that $$\int_0^4|f(x)-p(x)|^2dx<\epsilon.$$

It is very clear how to show this for $$\int_0^4|f(x)-p(x)|dx,$$ (which I've worked on here: Does integration preserve uniform convergence of sequence? (Weierstrass Approximation Theorem)) but I'm not sure how to deal with the square.

Answer 1

Note that if $|f(x)-p(x)|<\epsilon,$ then $|f(x)-p(x)|^2<\epsilon^2.$ So, it just comes down to fiddling with epsilons to remove the square. Spoiler below:

Fix $\epsilon>0.$ By WAT, there exists a polynomial $p$ so that $|f(x)-p(x)|<\sqrt{\epsilon}/2$ for all $x\in [0,4]$. Now, $$\int\limits_0^4|f(x)-p(x)|^2\, dx\leq \sup_{x\in [0,4]} |f(x)-p(x)|^2 \int\limits_0^4 \, dx<\epsilon.$$