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I want to show that if $(f_{n})_{n}$ converges weakly in $L^{p}(\Omega)$ then $(f_{n})_{n}$ is uniformly bounded in $L^{p}$.

The following is my attempt at proving this:

Assume $f_{n} \rightharpoonup 0$ in $L^{p}(\Omega)$ and let $\varphi \in (L^{p}(\Omega))^{*}$, then by Riesz Representation Theorem there exists a unique $u \in L^{p^{'}}(\Omega)$ such that:

$\langle \varphi, f_{n} \rangle = \int_{\Omega} u f_{n}$ for all $n \in \mathbb{N}$

Moreover, $||u||_{L^{p'}(\Omega)} = ||\varphi||_{(L^{p}(\Omega))^{*}}$

Since $f_{n} \rightharpoonup 0$ it follows that $\langle \varphi, f_{n} \rangle \rightarrow 0$ by characterization of weak convergence. We can also define the linear functional as a linear functional on $L^{p'}(\Omega)$ by $\langle \gamma_{n}, u \rangle := \langle \varphi, f_{n} \rangle$, it follows then that $\langle \gamma_{n},u \rangle \rightarrow 0$ is bounded for any $u$ since every convergent sequence is bounded.

$\therefore$ $\text{sup}_{n} |\langle \gamma_{n},u \rangle| < \infty$ by "Uniform Boundedness Principle" it follows that $\text{sup}_{n}||\gamma_{n}||_{(L^{p'})^{*}} < \infty$.

The result that I want is $\text{sup}_{n}||\gamma_{n}||_{(L^{p'})^{*}} = ||f_{n}||_{L^{p}}$. By Riesz Representation Theorem what I have is $||u||_{L^{p'}} = ||\varphi||_{(L^{p})^{*}}$, as stated above.

Can anyone see how this desired result follows from my argument? Is there a different, more efficient way of getting this result? Is this result unique to $L^{p}$ spaces? Thanks.

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    Just think of the $\gamma_n$ as the $f_n$ living in $L_p^{}$ (the action of $\gamma_n$ on $f\in L_p^*$ being pointwise evaluation). You have, for every $n$, $\Vert f_n\Vert = \Vert \gamma_n\Vert$ and that the $\gamma_n$ are pointwise bounded in $L_P^{}$. – David Mitra Jan 19 '14 at 15:44
  • I don't understand exactly what you mean. In my proof $\gamma_{n}$ is defined so that it acts on $u \in L^{p'}$. The result that I want to show is $\text{sup}{n}||\gamma{n}|| = ||f_{n}||$. Could you explain how it follows exactly from my proof? –  Jan 19 '14 at 16:02
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    First, there's no reason to introduce $\phi$ or $u$. I'm not sure I follow this part of your proof. You want to define $\gamma_n\in L_p^{*}$ by $\gamma_n(f^)=f^(f_n)$ for $f^\in L_P^*$. If you do this, then $\gamma_n$ is the image of $f_n$ under the canonical embedding of $L_p$ into its second dual. As such, you'll have the equality of the norms that you want. Also, your argument using the UBP will go through. – David Mitra Jan 19 '14 at 16:24
  • I agree, it's much better, thanks. –  Jan 20 '14 at 14:59

1 Answers1

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Expanding on a comment by David Mitra:

You want to define $\gamma_n\in L_p^{**}$ by $\gamma_n(f^*)=f^*(f_n)$ for $f^*\in L_p^*$. If you do this, then $\gamma_n$ is the image of $f_n$ under the canonical embedding of $L_p$ into its second dual. Since the family $\{\gamma_n\}$ is pointwise bounded on $L_p^*$, it is bounded in norm (by the Uniform Boundedness Principle). And since $\|\gamma_n\|_{L_p^{**}}=\|f_n\|_{L_p}$, the conclusion follows.

By the way, nothing here relies on the underlying space being $L_p$; the proof works the same in every normed space $X$. (The dual space $X^*$ is automatically complete, so the Uniform Boundedness Principle applies to it.)

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