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I want to go on the analysis of the sequence of functions $f_n(x)= e^{-\frac{x^2}{n^2}}$ on $\mathbb{R}$. Convergence in the sense of distributions and in $L^p$ can be found here: Convergence in $\mathcal{D}'(\mathbb{R})$; Convergence in $L^p(\Bbb R)$ Now the question is whether it converges weakly in $L^p(\mathbb{R})$ for $p\in [1,\infty)$. I would say no because $\sup_n \vert\vert f_n\vert\vert_{L^p(\mathbb{R})}=\infty$ as already mentioned in a last post. But how do we show this little lemma which says: $f_n \rightharpoonup f$ in $L^p(\mathbb{R})$ $\Leftrightarrow$ $\sup_n \vert\vert f_n\vert\vert_{L^p(\mathbb{R})}<\infty$ and $f_n\rightarrow f$ as distributions? For which $p$ is it true?

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Suppose $f_n \rightharpoonup f$ in $L^p(\mathbb{R})$, the boundedness of $f_n$ is an immediate consequence of the following more general statement: let $(E,\| \cdot \|_{E})$ be a normed vector space over $\mathbb{R}$ and $(x_n)_n \subset E,x \in E$, then $$ x_n \rightharpoonup x \implies \sup_{n} \| x_n \|_{E} < +\infty $$ i.e, every weakly convergent sequence is bounded. The proof is classical and relies on the canonical map properties (in particular, isometry), the uniform boundedness principle and the dual characterization of the norm.

Suppose $p \in (1,+\infty)$. As usual, if $f \in L^p$, we identify $f$ with its induced distribution $T_f : \mathcal{D}(\mathbb{R}) \mapsto \mathbb{R}$ such that $T_f(\phi)=\int_{\mathbb{R}} f\phi$ for every $\phi \in \mathcal{D}(\mathbb{R})$.

Recall that $f_n \rightharpoonup f$ in $L^p(\mathbb{R})$ is equivalent to say: $$ \int_{\mathbb{R}} f_n g \rightarrow \int_{\mathbb{R}} fg \qquad \forall g \in L^{p'}(\mathbb{R}) $$ On the other hand, $C^{\infty}_c (\mathbb{R}) \subset L^{p'}$, whence $f_n \rightarrow f $ in the sense of distributions.

We now prove the reverse implication. Again suppose $p \in (1,+\infty)$. Let $c=\max\{ \sup_n \| f_n \|_{p}, \| f \|_{p} \}<\infty $ and $p'$ be Holder conjugate of $p$. Let $g \in L^{p'}$, we want to prove that

$$ \int_{\mathbb{R}} (f_n-f)g \rightarrow 0 $$

Since $ (C_c^{\infty}(\mathbb{R}), \| \cdot \|_{p'})$ is dense in ($L^{p'},\| \cdot \|_{p'})$, for every $\epsilon > 0$ there exists a function $\rho \in C_c^{\infty}(\mathbb{R}) $ such that $ \|\rho-g\|_{p'} \le \frac{\epsilon}{2c}$ . For every $ n \in \mathbb{N}$ it holds:

\begin{align*} \| (f_n-f)g \|_{1} = \| (f_n-f)(g-\rho+\rho) \|_{1} \le \|(f_n-f)(g-\rho) \|_{1}+\| (f_n-f) \rho \|_{1} \\ \le \| (f_n-f) \rho \|_{1} + \| f_n - f \|_{p} \| g-\rho \|_{p'} \le \| (f_n-f) \rho \|_{1} + 2c \| g-\rho \|_{p'} \\ \le \| (f_n-f) \rho \|_{1} + \epsilon \end{align*}

Where we both used Holder's and Minkowski's inequalities. Since $\epsilon$ was arbitrary and $f_n \rightarrow f$ on test functions, the claim follows taking the limit as $ n \rightarrow +\infty $.

To conclude, let's see a counterexample for the right-left implication if $p=1$. Let $$ f_n(x)=\chi_{[n,n+1]} (x) \qquad x \in \mathbb{R} $$ We have $\sup_{n} \| f_n \|_{1} = 1 $ and $f_n \rightarrow 0 $ in the sense of distributions because for sufficiently large values of $n$, $f_n$ escapes the compact support of every test function. On the other hand the weak convergence in $L^1$ to the zero function would imply:

$$ \int_{\mathbb{R}} f_n g \rightarrow 0 \qquad \forall g \in L^{\infty}(\mathbb{R}) $$

Take $g=1$ to see that this is impossible. It's worth to note that the above proof in the case $p=1$ fails because test functions are not dense in $L^{\infty}$, hence the function $\rho$ that approximates $g$ may not exist.

As a final remark, note that "$ \leftarrow$" holds if $p=\infty$ replacing weak convergence with weak * convergence. Indeed, if $p \in (1,+\infty) $, the statement still trivially holds true in both directions since in these hypothesis $(L^p(\mathbb{R}, \| \cdot \|_{p})$ is reflexive and weak * convergence and weak convergence coincide in reflexive spaces.

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  • $\begingroup$ thanks a lot but this is only the direction $"\rightarrow"$, isn't it? the other direction would be more interesting for the problem above $\endgroup$ – tubmaster Mar 21 '17 at 16:29
  • $\begingroup$ Yes it is. I just had it immediately and so I wrote it down :) I think I can also handle the reverse implication: editing the answer as soon as I have time to work on it. $\endgroup$ – GaC Mar 21 '17 at 16:37
  • $\begingroup$ really, thanks a lot!! $\endgroup$ – tubmaster Mar 21 '17 at 16:55
  • $\begingroup$ Also added a counterexample for the right-left implication if $p=1$! $\endgroup$ – GaC Mar 22 '17 at 14:39
  • $\begingroup$ in your definition for $c$, do you mean min instead of max? $\endgroup$ – tubmaster Mar 22 '17 at 15:24

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