Does integration preserve uniform convergence of sequence? (Weierstrass Approximation Theorem)

Question

Trying to solve the following problem:

Let $f(x)$ be a continuous real-valued function on $[0,3]$. Given any $\varepsilon>0$ prove there exists a polynomial, $p(x)$, such that $\int_0^3|f(x)-p(x)|\,dx<\varepsilon$

This almost seems trivially true, which leads me to believe that I'm thinking about it incorrectly. If by Weierstrass theorem we know there exists a sequence of polynomials $P_n(x)$ in $[0,3]$ such that $\lim_{n \to \infty} P_n(x)=f(x)$, then if we set $p(x)=\lim_{n \to \infty} P_n(x)=f(x)$, then $|f(x)-p(x)|=|f(x)-f(x)|=0$ and therefore it is obviously true that $\int_0^3|f(x)-p(x)|\,dx<\varepsilon$. I'm almost certain this is not correct, so what am I doing wrong?

Answer 1

SW says that for any $\epsilon > 0$ there is a polynomial $p(x)$ such that $\forall x \in [0, 3]$ we have $|f(x) - p(x)| < \epsilon / 4$.

Then $$ \int_0^3 |f(x) - p(x)|\text{d}x \le (3 - 0) \times \sup_{x \in [0, 3]}|f(x) - p(x)| < 3 \times \frac{\epsilon}{4} < \epsilon $$

Answer 2

Your problem is that $p=\lim P_{n}(x)$ may not be a polynomial. I understand your idea and i fixed: By Weierstrass's theorem, there exists a sequence of polynomials such that $P_{n} \to f$ uniformly. So, given $\varepsilon>0$, there exists $n_{0}$ such that $|f(x)-P_{n_{0}}(x)|<\frac{\varepsilon}{6}$ for all $x \in [0,3]$ (notice that tis equivalent to say that $\sup_{x \in [0,3]} |f(x)-P_{n_{0}}|<\frac{\varepsilon}{6}$). So, $$\int_{0}^{3}|f(x)-P_{n_{0}}(x)|dx \leq \int_{0}^{3} \frac{\varepsilon}{6}dx=\frac{\varepsilon}{2}<\varepsilon.$$