$f,f_n \in C[-1,1]$, $f_n \rightrightarrows f \implies \lim_{n \to \infty} \Vert f_n-f \Vert_{1} = 0$

Question

Let $f,f_n \in C[-1,1]$ for $n \in \{1,2,3,\ldots\}$.

Prove if $f_n$ converges uniformly to $f$ on $[-1,1]$, then $\Vert f_n-f \|_{1} \to 0$ as $n \to \infty$.

My attempt:

Since $f_n \rightrightarrows f$, given an $\epsilon \gt 0$, $\exists N \in \mathbb{N}$, such that if $n \ge N$, $|f_n(x) - f(x)| \lt \epsilon, \ \forall x \in [-1,1]$. $$\implies \| f_n - f \|_{\infty} \lt \epsilon \tag{2}$$ As shown here, we know$\| f_n-f \|_{1} \le 2\| f_n-f \|_{\infty}$. So $\lim_{n \to \infty} \| f_n-f \|_1 = 0$.

I think the punchline was in recognizing $(2)$. I would like some suggestions if my proof is wrong, please.

Edit: I would also like to see an alternative proof if possible.

Answer 1

Your proof is correct, but I would like to point out that, from $|f_{n}(x)-f(x)|<\epsilon$, $x\in[-1,1]$, you can simply take integrals both sides to obtain $\displaystyle\int_{-1}^{1}|f_{n}(x)-f(x)|dx\leq\int_{-1}^{1}\epsilon dx=2\epsilon$, so $\|f_{n}-f\|_{1}\leq 2\epsilon$, this means $\|f_{n}-f\|_{1}\rightarrow 0$.