Show convergence in C-norm implies convergence in $L_p$-norm

Question

I attempted to start with the $L_p$ norm and raise it to the power of $p$ but got stuck because I realized that I have no idea how to eliminate the integrand.

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$L_p$ norm: $||f||_p = ||f||_{L_p[a,b]} = (\int_{a}^{b}~|f(x)|^p~~dx)^{\frac{1}{p}}$

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C-norm:

Answer 1

We have $|f_n(t)-f(t)| \le ||f_n-f||$ for all $f,f_n \in C$ , all $n \in \mathbb N$ and all $t \in [a,b]$. Hence

$|f_n(t)-f(t)|^p \le ||f_n-f||^p$ for all $f,f_n \in C$ , all $n \in \mathbb N$ and all $t \in [a,b]$.

This gives

$ \int_a^b|f_n(t)-f(t)|^p dt \le \int_a^b ||f_n-f||^p dt =(b-a)||f_n-f||^p$.

Therefore

$||f_n-f||_p \le (b-a)^{1/p}||f_n-f||$.

Conclusion: $||f_n-f|| \to 0$ implies $||f_n-f||_p \to 0$ .

Answer 2

Note that $(\int _a^b|f_n-f|^p)\le \max_{x\in [a,b]} |f_n(x)-f(x)|^p\times (b-a)^p$

Choose $\epsilon<1$.

Now since $f_n\to f$ in $C-$ Norm

$\implies ||f_n-f||=\max_{x\in [a,b]} |f_n(x)-f(x)|<\epsilon(<1)\implies |f_n(x)-f(x)|<\epsilon$

$\implies |f_n(x)-f(x)|^p<\epsilon$

$\implies \max_{x\in [a,b]} |f_n(x)-f(x)|^p< \epsilon$ for arbitrary $\epsilon>0$.

Hence $||f_n-f||_p=(\int _a^b|f_n-f|^p)^{\frac{1}{p}}< \epsilon (b-a)$