Showing Convergence in $L^p$ norms

Question

Let $X$ be a finite measure space and $1\le p<\infty$ and $\{f_n\}$ be a sequence in $L^p(X)$ such that coverge to $f$ in $L^p(X)$ . If there exists constant $K$ such that for every $n\in \mathbb{N}$ and every $x\in X$ $$|f_n(x)|\le K $$ then for every $1\le r<\infty$ we have $$||f_n-f||_{r} \to 0 $$ which $|| \ .||_p $ is $ L^p $ norm.

Answer 1

Suppose the conclusion is not true, so there $\exists r, \geq 1$ $\epsilon >0$ and a subsequence $\{f_{n_k}\}$ such that $\|f_{n_k} - f\|_r \geq \epsilon$.

But $f_{n_k}$ converges in $L^p$ to $f$, so its has a sub-subsquence $\{f_{n_{k_l}}\}$ such that $\{f_{n_{k_l}}\}$ converges to $f$ almost surely.

Then by dominated convergence, we get $f_{n_{k_l}}$ conveges in $L^r$ to $f$, which is contradictory with $\|f_{n_k} - f\|_r \geq \epsilon$.

So it's done.

Answer 2

Since $f_n\to f$ in $L^p$ and $X$ has finite measure, then $f_n\to f$ in $L^q$ for any $1\leq q\leq p$.

For $p<r<+\infty$, you know that $|f_n(x)|\leq K$ uniformly. Convergence in $L^p$ gives convergence a.e. along subsequence $f_{n_k}\to f$, and by dominated convergence $f_{n_k}\to f$ in $L^r$. If there exists another sequence $f_{m_k}\to g\neq f$ in $L^r$, then (since $r>p$) you have $f_{m_r}\to g\neq f$ in $L^p$, contradicting the hypothesis $f_n\to f$ in $L^p$.