How do I prove that $f_n\to f$ in $L^p$?

Question

Let $\{f_n\}$ be a sequence in $L^p([0,1])$ for $p\geq 1$. Suppose there exists $f\in L^p([0,1])$ satisfying $\lim_{n\to\infty} \int_0^1 f_n(x)g(x)dx = \int_0^1 f(x)g(x)dx$ for any $g\in L^2([0,1])$. Moreover, assume that $\lim_{n\to\infty} ||f_n||_p=||f||_p$. In this case, how do I prove that $f_n\rightarrow f$ in $L^p$?

Even when $p=2$, the result is non-trivial since $\lim_{n\to\infty}<f_n-f,g>=0$ (weak convergence) does not simply imply the convergence in $L^2$.

More seriously, if $p\neq 2$, I'm not sure what's going on here. Why for given $f\in L^p$ and $g\in L^2$, $fg\in L^1$?. Holder inequality cannot be applied here.

Moreover, I'm not sure how to use the condition $\lim_{n\to\infty} ||f_n||_p =||f||_p$. If $f_n \to f$ pointwise a.e., this condition seems useful, but this is not the case. How do I prove this? Thank you in advance.

In $L^2$, the result is quite straightforward: Just write $|f-f_n|_2^2=\langle f-f_n,f-f_n\rangle$ and go to town expanding the r.h.s. using linearity. — Alex R., May 26 '16 at 03:24
Are you sure that $g\in L^2$? It would be much simpler if $g\in L^q$ where $\frac 1q+\frac1p=1$, as you must have noticed. — BigbearZzz, May 26 '16 at 03:48
@BigbearZzz Yes, I'm sure it is $2$, not the conjugate exponent unless there is a typo. This problem was on qualifying exam. — Rubertos, May 26 '16 at 03:50
@Rubertos I've manged to prove it for the case when $p\in [1,2]$. For $p>2$ I still cannot think of a way. :( — BigbearZzz, May 26 '16 at 05:15
I'm curious how you proved it for the case $1\leq p \leq 2$. If $p>2$, then by setting $E={|f|\geq 1}$, $\int |fg|=\int_E |fg|+ \int_{E^c} |fg| \leq \int_E |f|^{p/2}|g| + \int |g| \leq (\int |f|^p)(\int |g|^2) + \int |g| <\infty$. Hence, $fg$ is of $L^1$. This is all I know so far.. — Rubertos, May 26 '16 at 05:22

score 2 · Answer 1 · edited Apr 13 '17 at 12:20

2

Disclaimer: This answer is incomplete, I've only managed to prove the case where $p\in (1,2]$. The cases $p>2$ and $p=1$ remains unknown to me. Please feel free to criticize my answer if I made any mistakes.

Let $p\in (1,2]$ and $q$ be its conjugate satisfying $\frac 1p + \frac 1q = 1$. Since $I=[0,1]$ is a set of finite measure, the $L^p[0,1]$ spaces are downward directed in the sense that $q<q'$ implies $$ L^{q'}\subset L^q $$ for $q,q'\ge 1$. By the assumption that $p\in (1,2]$, we have $q\ge 2$ so $L^2$ includes all functions in $L^q$.

By duality, we have $f_n\rightharpoonup f$ since $\{g\}$ include all function in $L^q$. Together with $||f_n||_p\to ||f||_p$ we have $||f_n-f||_p\to 0$. Consult this page for more information.

edited Apr 13 '17 at 12:20

Community

1

answered May 26 '16 at 05:29

BigbearZzz

15,084

Not sure why your proof works for the case $p=1$. – Rubertos May 26 '16 at 05:34
@Rubertos Good question, actually I didn't. I miss-typed it. :P – BigbearZzz May 26 '16 at 05:38
I posted an answer proving that $f_n\to f$ weakly, and referred to the same question/page as you. – Luiz Cordeiro May 26 '16 at 05:47
$L^2$ is dense in $L^p$ for every $p$, since it contains $C^\infty_c$. hence $f_n \to f$ weakly in $L^p$. and together with $|f_n|_p \to |f|_p$ we get $|f_n-f|_p \to 0$ – reuns May 26 '16 at 06:27
@user1952009 That is indeed interesting! Thank you very much. I'm still new to functional analysis so there's a lot for me to learn. – BigbearZzz May 26 '16 at 11:23

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

This solves the case $p>1$: Let's show that $\int f_ng\to\int fg$ for all $g\in L^q=L^q[0,1]$.

Let $g\in L^q$. Let's show that $\int f_ng$ is Cauchy. Let $\epsilon>0$. Choose $p\in C[0,1]$ with $\Vert g-p\Vert_q<\epsilon$. Then for $n,m$ large, $$|\int f_ng-\int f_mg|\leq|\int f_n(g-p)|+|\int f_np-f_mp|+|\int f_m(p-g)|$$ Applying Hölder's inequality in the first and the third terms in the RHS, and using the fact that the sequence $(\Vert f_n\Vert_p)$ is bounded, say $K=\sup_n\Vert f_n\Vert_p$, yields $$|\int f_ng-\int f_mg|\leq 2\epsilon K+|\int f_np-f_mp|\tag{$*$}$$ which is small for $n,m$ large (because $K$ does not depend on $\epsilon$).

Let $L=\lim_n\int f_ng$. We will show that $L=\int fg$. Indeed, take $\epsilon$, $p$ and $K$ as above, and note that \begin{align*} |L-\int fg|&\leq|L-\int f_ng|+|\int f_n(g-p)|+|\int f_np-\int fp|+|\int f(p-g)|\\ &\leq|L-\int f_n g|+K\epsilon+|\int f_np-\int fp|+K\epsilon \end{align*} and both $|L-\int f_n g|$ and $|\int f_np-\int fp|$ go to $0$. This shows that $|L-\int fg|<2K\epsilon$ for all $\epsilon$, and therefore $L=\int fg$.

Therefore, $f_n\to f$ weakly. Refering to the same question as in BigbearZzz's answer, we conclude that $f_n\to f$ in $L^p$.

This is much better that my answer, definitely +1 to this. – BigbearZzz May 26 '16 at 11:28 — BigbearZzz, May 26 '16 at 11:28

How do I prove that $f_n\to f$ in $L^p$?

2 Answers2