# A characterization of Weak Convergence in $L^p$ spaces

I'm working on the following problem, I'm having trouble with the reverse direction. My question is bolded below. Also could someone check my forward direction?:

Let $$(X, \mathcal{M}, \mu)$$ be a $$\sigma$$ finite measure space and $$\{f_n\},f \in L^P(X)$$. Prove that $$f_n \rightharpoonup f$$ in $$L^p(X)$$ iff $$\|f_n\|_p \leq c$$ for all $$n$$ and $$\int_A f_n\, d\mu \rightarrow \int_A f \, d\mu$$ for all $$A$$ with $$\mu(A) < \infty$$.

For the reverse direction, we can use the characteristic functions in $$L^q$$ to build arbitrary functions in $$L^q$$ and use Monotone Convergence on $$A$$ equals a ball. Then increase the radius of the ball at each step making error $$\epsilon/2^n$$. However, I'm having trouble seeing how I use the boundedness of the sequence $$f_n$$)

(For the forward direction, choosing $$\chi_{A}\in L^q(X)$$ will get the integral condition and the $$\|f_n\|_p$$ were bounded because the sequence originally lived in $$L^p(X)$$.)

• For the "for the forward direction", i don't see why the fact that the $f_n$'s live in $L^p$ means that their norms are uniformly bounded. – mathworker21 May 27 '19 at 13:29
• Was about to say the same thing as @mathworker21. For example, $X=[0,1],$ $f_n\equiv n.$ – zugzug May 27 '19 at 13:32
• @zugzug I mean, yea, any time you have a norm, you can, by definition, linearly multiplicatively scale to arbitrarily increase the norm – mathworker21 May 27 '19 at 13:34
• You need to use the fact that the sequence converges weakly, combined with an application of the uniform boundedness principle to see that $\sup_n \|f_n\|_{L^p} < \infty$ in the forward direction. As has already pointed out, the fact that $f_n \in L^p$ for each $n$ is not enough. – Rhys Steele May 27 '19 at 13:48
• @RhysSteele Can you elaborate on this comment? zugzug posted a longer explanation but I had questions about their explanation as well – yoshi May 29 '19 at 0:40

Here is how the boundedness of the sequence $$\{f_n\}$$ should enter your argument in the reverse direction:

Let $$g\in L^q$$. You want to show that $$\int f_n g\, d\mu \rightarrow \int f g \,d\mu$$. To this end, construct a sequence of simple functions $$g_m$$ such that $$g_m\rightarrow g$$ strongly in $$L^q$$ (I guess this is what you mean by "build"). You can then deduce

$$\begin{split} \left|\int (f_n-f) g\, d\mu \right| &\leq \left|\int (f_n-f) (g-g_m)\, d\mu \right| + \left|\int (f_n-f) g_m\, d\mu\right|\\ &\leq \|f_n-f\|_p\|g-g_m\|_q + \left|\int (f_n-f) g_m\, d\mu\right| \\ &\leq (\|f_n\|_p+\|f\|_p)\|g-g_m\|_q + \left|\int (f_n-f) g_m\, d\mu\right| \\ &\leq (C+\|f\|_p)\|g-g_m\|_q+ \left|\int (f_n-f) g_m\, d\mu\right| <\epsilon \end{split}$$

for $$n$$ sufficiently large, provided $$\|f_n\|_p$$ is bounded by $$C$$. More precisely, you first choose $$m$$ sufficiently large so that the first term above is less than, say, $$\frac{\epsilon}{2}$$, then choose $$n$$ sufficiently large so that the second term above is less than $$\frac{\epsilon}{2}$$.

P.S. Your should include your assumptions on $$p$$. I am assuming $$p\in(1,\infty)$$.

• ah, okay so is it correct to say that following? -- the uniform bound on $\|f_n\|$ is for the $\|f_n-f\|_p\|g-g_m\|$ piece (it ensures the first term is bounded) and the $\int_A f_n \rightarrow \int_A f$ conditions ensures the second term converges. – yoshi May 28 '19 at 14:57
• @yoshi: Yes, exactly like that. – StarBug May 28 '19 at 17:01

This answer will only deal with the forward implication since you have a good answer dealing with the other one.

First, let's show that weak convergence implies boundedness of norms. This is in fact true in Banach spaces in general by essentially the same idea but I will work in this special case.

Recall that $$L^p(X)^*$$ is isometrically isomorphic to $$L^q(X)$$ where $$p^{-1} + q^{-1} = 1$$. In particular, we can identify each $$f_n$$ with a linear functional $$\phi_n$$ on $$L^q$$ defined by $$\phi_n(g) = \int_X f_n g d\mu$$ and have $$\|f_n\| = \|\phi_n\|$$.

Now we seek to apply the uniform boundedness theorem to the family $$\{\phi_n\}_{n \geq 1}$$. To do this, notice that for fixed $$g \in L^q(X)$$, $$\phi_n(g) \to \int_X fg d\mu$$ by weak convergence of $$f_n$$. Since convergent sequences in $$\mathbb{R}$$ (or $$\mathbb{C}$$) are bounded this implies that $$|\phi_n(g)|$$ is a bounded sequence for each $$g \in L^q(X)$$.

In turn, by the Uniform Boundedness Theorem, we have that $$\sup_n \|f_n\| = \sup_n \|\phi_n\| < \infty$$.

The second part is immediate since $$1_A \in L^q(X)$$ so that $$\int_A f_n d \mu = \int_X 1_A f_n d \mu \to \int_X 1_A f d \mu = \int_A f d \mu$$ by the weak convergence.

• this was clear, thanks! I upvoted, I'll award some bounty in liu of accepting the answer, when the option becomes available. – yoshi May 29 '19 at 12:24

For the forward direction:

Define $$T_n(g):=\int_X f_n g$$ to be a family of linear operators on $$L^q$$. Also, define $$T(g):=\int_X f g$$. Then for each $$g,$$

$$|(T_n-T)(g)|\leq ||f_n-f||_p ||g||_q Note this bound depends on $$g,$$ but not $$n$$ because $$f_n$$ converges to $$f$$ in $$L^p.$$ By the uniform boundedness theorem, the dependence on $$g$$ can go away and so $$||T_n-T|| \to 0$$ in operator norm. By duality, $$||T||=||f||_p$$ and $$||T_n||=||f_n||_p.$$ Therefore, for large $$n,$$ we have $$||T_n||<||T||+\epsilon$$. This is equivalent to $$||f_n||_p < ||f||_p + \epsilon.$$ The second part is trivial by Holder's inequality.

• second part is trivial by Holder's?? You mean it's trivial because $1_A \in L^q$...? – mathworker21 May 28 '19 at 1:08
• thanks for this clarification! alas, I also got the forward direction wrong then too. Some questions: 1. In more detail: "$g$ dependence goes away" because the bounded linear operator $T_n-T$ is pointwise bounded on $L^q$, so by unf bdd princ its uniformly bdd. Thus $\sup_n\|T_n - T\| < \infty$. Separately, I have assumed weak convergence so $\int f_n g \rightarrow \int f g$ for all $g \in L^q$. This implies $|(T_n - T)(g)| = |\int (f_n - f) g| < \epsilon$ for all $g \in L^q$. This seems to imply the result obtained using unf bdd princ, so why was unf bdd necessary? – yoshi May 28 '19 at 14:45
• 2. For the dual norm, I'm obtaining $\|T\| \leq \|f\|_p$ using holder. How do you obtain $\|T\| \geq \|f\|_p$? – yoshi May 28 '19 at 14:46
• There was a typo in my first question. Second Try: 1. In more detail: "$g$ dependence goes away" because the bounded linear operator $Tn−T$ is pointwise bounded on Lq, so by unf bdd princ its uniformly bdd WE HAVE $\sup_n\|T_n−T\| \leq \infty$. (The rest is the same) – yoshi May 28 '19 at 15:07
• The answer for the first part is incorrect. It is true that the family of operators $T_n$ is uniformly pointwise bounded but not by the attempt here since $f_n$ only converges weakly to $f$ and so you can't just assume a uniform bound on $\|f_n - f\|$ (it's what you're trying to prove!) – Rhys Steele May 29 '19 at 7:41