Let $\{f_n\}\subset L^2([0,1])$ be such that $f_n\to f$ pointwise and $\sup \|f_n\|_{L^2}^2 < \infty$, show that $f_n\to f$ in $L^p$ for $p\in [1,2)$.

Question

I wish to prove the following statement: Suppose $\{ f_n\} \in L^2([0,1])$, $f_n \to f$ pointwise, and $\sup \|f_n\|_{L^2}^2 < \infty$. Then $f_n \to f$ in $L^p$ for $p\in [1,2)$.

Here is what I have so far: By Fatou's lemma, one can see that $$ \int_{[0,1]} |f|^2 \ dx \leq \liminf \int_{[0,1]} |f_n|^2 \ dx < \infty $$ so indeed $f \in L^2$ as well. However, after this step, I am stuck. I tried applying Holder's inequality to yield $$\int_{[0,1]} |f_n-f|^p \ dx \leq \left(\int_{[0,1]} |f_n-f|^2 \ dx\right)^{p/2} $$ But this dosent seem helpful as we do not have $f_n \to f$ in $L^2$. I would appreciate any help on this problem.

Edit: After thinking for a while, I came up with the following idea using Dominated Convergence Thm. We have the bound $$\|f_n - f\|_{L^p} \leq \|f_n\|_{L^p} + \|f\|_{L^p}$$ The RHS is integrable using the Holder inequality as above. The result would then follow directly by passing the limit inside the integral. Would this approach work?

Answer 1

Feng as already explained why the dominated convergence argument does not work.

Here is an alternative proof, using Egoroff's theorem. It was shown in the opening post that $f$ belongs to $\mathbb L^2$, hence considering $g_n=f_n-f$, we know that $g_n\to 0$ pointwise and that $M:=\sup_{n\geqslant 1}\lVert g_n\rVert_2 <\infty$. We have to show that for each $p\in [1,2)$, $\lVert g_n\rVert_p\to 0$.

Let $\varepsilon>0$. Egoroff's theorem that we can find a set $A_\varepsilon\subset [0,1]$ such that $\sup_{x\in A_\varepsilon}\lvert g_n(x)\rvert\to 0$ and $\lambda\left([0,1]\setminus A_\varepsilon\right)<\varepsilon$. Consequently, $$ \lVert g_n\rVert_p^p=\int_{A_\varepsilon} \lvert g_n(x)\rvert^p d\lambda(x)+\int_{[0,1]\setminus A_\varepsilon} \lvert g_n(x)\rvert^p d\lambda(x)\leqslant \sup_{x\in A_\varepsilon}\lvert g_n(x)\rvert+\int_{[0,1]\setminus A_\varepsilon} \lvert g_n(x)\rvert^p d\lambda(x).$$ For the last integral, use Hölder's inequality with the exponents $p'=2/p$ and $q'=2/(2-p)$ to get that $$ \lVert g_n\rVert_p^p\leqslant \sup_{x\in A_\varepsilon}\lvert g_n(x)\rvert+M^p\varepsilon^{(2-p)/2}. $$

Answer 2

Your approach using DCT is wrong. To show that $\|f_n-f\|_{L^p}\to0$ directly using DCT, you should find a $g\in L^p$ such that $|f_n-f|\leq g$ a.e. In your post, you only get $\sup_{n}\|f_n-f\|_{L^p}<+\infty$, which is NOT sufficient to conclude $\|f_n-f\|_{L^p}\to0$.

Below is my proof.

A well-known identity. Let $f:[0,1]\to\mathbb R$ be a measurable function. For each $\lambda>0$, define $$E_\lambda:=\{x\in[0,1]: |f(x)|>\lambda\},$$ then $$\|f\|_{L^p([0,1])}=p\int_0^\infty \lambda^{p-1}m(E_\lambda)\,d\lambda,\qquad \forall p>0.$$ Indeed, Fubini's theorem implies that \begin{align*} \|f\|_{L^p([0,1])}&=\int_{[0,1]}|f(x)|^p\,dx=p\int_{[0,1]}\left(\int_{[0,|f(x)|]}\lambda^{p-1}\,d\lambda\right)\,dx\\ &=p\int_{[0,1]}\int_0^\infty \chi_{[0,|f(x)|]}(\lambda)\lambda^{p-1}\,d\lambda dx\\ &=p\int_0^\infty \lambda^{p-1}\left(\int_{[0,1]}\chi_{[0,|f(x)|]}(\lambda)\,dx\right)\,d\lambda\\ &=p\int_0^\infty \lambda^{p-1}\left(\int_{[0,1]}\chi_{E_\lambda}(x)\,dx\right)\,d\lambda\\ &=p\int_0^\infty \lambda^{p-1}m(E_\lambda)\,d\lambda. \end{align*}

Now let $\{ f_n\} \subset L^2([0,1])$ be such that $f_n \to f$ pointwise and $\sup \|f_n\|_{L^2}^2 < \infty$. By Fatou, $f\in L^2$. Hence we can assume without loss of generality that $f\equiv0$, otherwise we consider $g_n=f_n-f$. For any $\lambda>0$ and $n\in\mathbb N$, we define $$E_{n,\lambda}:=\{x\in[0,1]: |f_n(x)|>\lambda\}.$$ Since $f_n\to0$ pointwise, we have $$\bigcap_{k=1}^\infty\bigcup_{n=k}^\infty E_{n,\lambda}=\emptyset, $$ then $\lim_{k\to\infty}m\left(\bigcup_{n=k}^\infty E_{n,\lambda}\right)=0$, and hence $$\lim_{n\to\infty}m(E_{n,\lambda})=0,\qquad \forall \lambda>0.\tag{$*$}$$ It follows from $\sup \|f_n\|_{L^2}^2 < \infty$ that $$\sup_{n\in\mathbb N}\int_0^\infty \lambda m(E_{n,\lambda})\,d\lambda\leq M$$ for some $M\in(0,+\infty)$.

Now we're ready to prove $\lim_{n\to\infty}\|f_n\|_{L^p}=0$ for $p\in[1,2)$. For any $\varepsilon>0$, there exists $R>0$ such that $R^{p-2}<\varepsilon$, then \begin{align*} \|f_n\|_{L^p}^p&=p\int_0^\infty \lambda^{p-1}m(E_{n,\lambda})\,d\lambda\\ &=p\int_0^R\lambda^{p-1}m(E_{n,\lambda})\,d\lambda+p\int_R^\infty\lambda^{p-1}m(E_{n,\lambda})\,d\lambda\\ &\leq p\int_0^R\lambda^{p-1}m(E_{n,\lambda})\,d\lambda+pR^{p-2}\int_R^\infty\lambda m(E_{n,\lambda})\,d\lambda\\ &\leq p\int_0^R\lambda^{p-1}m(E_{n,\lambda})\,d\lambda+pM\varepsilon; \end{align*} By $(*)$, $m(E_{n,\lambda})\leq 1$ and the dominated convergence theorem we have $$\lim_{n\to\infty}\int_0^R\lambda^{p-1}m(E_{n,\lambda})\,d\lambda=0,$$ hence $$\limsup_{n\to\infty}\|f_n\|_{L^p}^p\leq pM\varepsilon,\qquad \forall \varepsilon>0,$$ which implies that $\lim_{n\to\infty}\|f_n\|_{L^p}=0$ for $p\in[1,2)$.