Does $\|f_j-f\|_p \rightarrow 0$ imply $\|f_j\|_p \rightarrow \|f\|_p$?

Question

Let $f_j, f\in L^p(\Omega)$ such that $\|f_j-f\|_p \rightarrow 0 $ as $j\rightarrow \infty$. Does this imply that $\|f_j\|_p \rightarrow \|f\|_p$?

I know that there exists a subsequence $\{f_{j_k}\}\subset \{f_j\}$ such that $$ f_{j_k}(x) \rightarrow f(x) \quad \text{ for a.e. }x\in \Omega \quad \text{ as } \quad j_k\rightarrow \infty.$$ Does this result play any role in this?

Answer 1

Note that for $p\geq 1$: $$ \left| \|f_i\|_p-\|f\|_p \right| \leq \|f_i-f\|_p. $$ Therefore: $$ \lim_{n\to\infty} (\|f_i\|_p-\|f\|_p)=0 $$ and from $f_i,f\in L^p$, we have: $$ \lim_{n\to\infty} \|f_i\|_p= \|f\|_p. $$ When $p=1$, this is Scheffe's lemma.

Answer 2

With the inverse triangle equality you get that

$$\vert \Vert f_j\Vert_p - \Vert f\Vert_p\vert \leq \Vert f_j-f\Vert_p \to 0 \quad \text{as} \quad j \to \infty.$$

This works in general normed spaces (and is no special property of the $L^p$-spaces).

Answer 3

As Yaddle has pointed out, the property is true in any normed space. Abstract version of the solution: by the reverse triangle equality, the norm is continuous. Now, $$ \lim_{n\to\infty}f_n = f\implies \lim_{n\to\infty}||f_n|| = ||\lim_{n\to\infty}f_n|| = ||f||. $$