What Is an Example of Weak but Not Strong Convergence in L²(R)?

[dextercioby]

So I've seen the distinction one makes in case of infinite-dimensional Hilbert spaces. Weak convergence versus strong convergence of sequences.

I cannot think of an example of sequence of vectors in L^2(R) which converges with respect to the scalar product, but not with respect to the norm induced by it.

Can one offer me such an example ?

Another question would be: if the strong convergence induces the metric topology on L^2(R), then does weak convergence induce a topology ?

Thank you!

 

[micromass]

So I've seen the distinction one makes in case of infinite-dimensional Hilbert spaces. Weak convergence versus strong convergence of sequences.

Allright, let's first agree on the meaning here. I assume with strong convergence you mean:

x_n\rightarrow x~\text{strongly}~\Leftrightarrow~\|x_n-x\|\rightarrow 0

And with weak convergence:

x_n\rightarrow x~\text{weakly}~\Leftrightarrow~<x_n,y>\rightarrow <x,y>, \forall y.

Is this what you mean?

I cannot think of an example of sequence of vectors in L^2(R) which converges with respect to the scalar product, but not with respect to the norm induced by it.

Can one offer me such an example ?

I can offer you an easy example in \ell^2, is that also good? (it should be since L^2(R) is isometric to \ell^2)...

Let

x_1=(1,0,0,...),~x_2=(0,1,0,0,...),~x_3=(0,0,1,0,...),...

This sequence does not converge strongly, but it does converge weakly in \ell^2 (if I didn't make a mistake). I got this example from the Banach-Alaoglu theorem, which has as corollary for Hilbert spaces that "every bounded sequence in a Hilbert space has a weakly convergent subsequence".

Another question would be: if the strong convergence induces the metric topology on L^2(R), then does weak convergence induce a topology ?

Yes! It is called the weak topology. The weak topology has a lot of good properties that the strong topology doesn't have. For example, the closed unit ball in a Hilbert space has a weak compact closure is a nice result for the weak topology which does not hold for the strong topology. My example is again an incarnation of the Banach-Alaoglu theorem...

 

[dextercioby]

Allright, let's first agree on the meaning here. I assume with strong convergence you mean:

x_n\rightarrow x~\text{strongly}~\Leftrightarrow~\|x_n-x\|\rightarrow 0

And with weak convergence:

x_n\rightarrow x~\text{weakly}~\Leftrightarrow~<x_n,y>\rightarrow <x,y>, \forall y.

Is this what you mean?

Yes, these are the standard definitions.

I can offer you an easy example in \ell^2, is that also good? (it should be since L^2(R) is isometric to \ell^2)...

Let

x_1=(1,0,0,...),~x_2=(0,1,0,0,...),~x_3=(0,0,1,0,...),...

This sequence does not converge strongly, but it does converge weakly in \ell^2 (if I didn't make a mistake). I got this example from the Banach-Alaoglu theorem, which has as corollary for Hilbert spaces that "every bounded sequence in a Hilbert space has a weakly convergent subsequence".

OK, thank you for the example. I've done some research based on your hint and the inequivalence betweem the 2 convergence types is valid generally for any orthonormal set of vectors in a pre-Hilbert space.

Yes! It is called the weak topology. The weak topology has a lot of good properties that the strong topology doesn't have. For example, the closed unit ball in a Hilbert space has a weak compact closure is a nice result for the weak topology which does not hold for the strong topology. My example is again an incarnation of the Banach-Alaoglu theorem...

Ok, but how is the topology defined ? The metric topology in a pre-Banach space has the unit balls, what are the open sets that define the topology in the case of convergence of arbitrary sequences with respect to the scalar product ?

Thanks!

 

[micromass]

Ok, but how is the topology defined ?

Well, you take the weakest topology on H such that all the functions &lt;\cdot,y&gt;<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />\rightarrow \mathbb{C} are continuous. This topology is generated by the following subbasis:

\{&lt;\cdot,y&gt;^{-1}(G)~\vert~y\in H,~G~\text{open in}~\mathbb{C}\}

In topology, this is also known as the initial topology.

More generally, we can define the weak topology on a Banach spaces X (we can even do more general than that!) as the weakest topology such that all the functions in \mathcal{C}(X,\mathcal{C}) (= the continuous functions for the strong topology) are continuous. So, in particular, the weak and strong topology have the same continuous functionals.
Weak convergence for Banach spaces is defined as

x_n\rightarrow x~\Leftrightarrow~f(x_n)\rightarrow f(x)~\text{for}~f\in \mathcal{C}(X,\mathbb{C}).

Note that these definitions correspond to the usual ones in Hilbert spaces by the Riesz lemma: every continuous functional has the form &lt;\cdot,y&gt;.

 

[dextercioby]

Well, you take the weakest topology on H such that all the functions &lt;\cdot,y&gt;<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />\rightarrow \mathbb{C} are continuous. This topology is generated by the following subbasis:

\{&lt;\cdot,y&gt;^{-1}(G)~\vert~y\in H,~G~\text{open in}~\mathbb{C}\}

In topology, this is also known as the initial topology.

OK, that term I saw in an article by J.E. Roberts on Rigged Hilbert Spaces.

More generally, we can define the weak topology on a Banach spaces X (we can even do more general than that!) as the weakest topology such that all the functions in \mathcal{C}(X,\mathcal{C}) (= the continuous functions for the strong topology) are continuous. So, in particular, the weak and strong topology have the same continuous functionals.
Weak convergence for Banach spaces is defined as

x_n\rightarrow x~\Leftrightarrow~f(x_n)\rightarrow f(x)~\text{for}~f\in \mathcal{C}(X,\mathbb{C}).

So I'm supposed to read this: x_n converges weakly to x iff the the sequence of functionals's values converges weakly in C.

Thank you for the valuable input.