# How is the uniform boundedness principle compatible with this seemingly weak convergent sequence?

In showing that $(x_i\rightharpoonup x)\not\Rightarrow(x_i\to x)$ or similar noncorallaries, one frequently uses the counterexample $$(u_i)_{i\in\mathbb{N}}\in \ell^2\colon \quad u_i = (\underbrace{0,\ldots,0}_{i-1},1,0,\ldots)$$ This, if I can trust my professor, is a weak convergent sequence in $\ell^2$, because it is bounded (I apparently missed that bit, this was the cause of the confusion) and for each $\varphi_j=(0,\ldots,1,0,\ldots)$, there exists an $N\in\mathbb{N}$, namely $N=j$, such that for every $i>N$, $$\langle u_i, \varphi_j \rangle_{\ell^2} = 0 + \ldots + 0\cdot1 + 0 + \ldots + 1\cdot 0 + 0\ldots = 0 < \varepsilon$$ for all $\varepsilon>0$. Because the $\varphi_j$ form a complete orthonormal system in $\ell^2$, this is sufficient for $(u_i)_i$ to be weakly convergent.

The problem is now: consider the sequence $$(v_i)_{i\in\mathbb{N}}\in \ell^2\colon \quad v_i = (\underbrace{0,\ldots,0}_{i-1},i,0,\ldots).$$ You could show that this is weakly convergent in exactly the same way as I just did for $(u_i)_i$, but on the other hand, $(v_i)_i$ is obviously an unbounded sequence and according to the principle of uniform boundedness (which I don't quite understand) every weak convergent sequence is bounded.

So what's wrong here?

Well, as the answers pointed out that it was the requirement of $(u_i)$ being bounded which was necessary for using just the $\varphi_j$ to show weak convergence.

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I must say that I don't like the way you phrase your professor's proof at all, as there is a crucial ingredient missing (either this was a major mistake on your professor's side, or you forgot something). Namely boundedness of the sequence $(u_i)$.

First recall Bessel's inequality: For an orthonormal system $(u_{i})$ and all $x \in H$ the inequality $$\sum_{i} |\langle x, u_{i} \rangle|^2 \leq \Vert x \Vert^2$$ holds. This implies that we must have $\langle x, u_i \rangle \to 0$ for all $x$, and hence an orthonormal system converges weakly to zero. (that's my preferred way of showing that an orthonormal system converges weakly to zero). Note also that an orthonormal system is bounded, as $\|u_{i}\| = 1$.

Now there is the following result (which I guess was what your professor was referring to):

A sequence $(u_{i})$ converges weakly to zero if and only if it is bounded and there exists an orthonormal basis $(\phi_{n})$ such that $\langle u_{i}, \phi_{n} \rangle \to 0$ as $i \to \infty$ for all $n$.

Indeed, if $u_{i}$ converges weakly to zero then the condition is clearly fulfilled for any orthonormal basis by Bessel's inequality and the sequence is bounded by the uniform boundedness principle.

Conversely, assume that $\Vert u_{i} \Vert \lt C$ for all $i$ and assume that there exists an orthonormal basis such that $\langle u_{i}, \phi_{n} \rangle \to 0$ as $i \to \infty$ for all $n$. Fix $\varepsilon \gt 0$. Bessel's inequality tells us that for every $x \in H$ there exists $N$ such that $\langle x, \phi_{n} \rangle \leq \varepsilon$ for all $n \geq N$. Choose $i$ so large that $\langle \phi_{k}, u_{i} \rangle \leq \varepsilon$ for all $k \lt N$. Then we can estimate $$|\langle u_{i}, x \rangle| = \left\vert \langle u_{i}, \sum_{n} \langle x_{i}, \phi_{n} \rangle \phi_{n} \rangle \right\vert\leq \sum_{k \lt N} \underbrace{|\langle u_{i}, \phi_{k}\rangle|}_{\leq \varepsilon}\, \underbrace{|\langle x, \phi_{k} \rangle|}_{\leq \|x\|} + \sum_{n \geq N} \underbrace{|\langle u_{i}, \phi_{n}\rangle|}_{\lt C}\, \underbrace{|\langle x, \phi_{n} \rangle|}_{\leq \varepsilon}$$ so that $|\langle u_{i}, x \rangle| \lt (\|x\| + C)\varepsilon$ for all large enough $i$. As $\varepsilon \gt 0$ was arbitrary this means that $|\langle u_{i}, x \rangle| \to 0$ for all $x$, and hence $u_{i}$ converges weakly to zero.

As Luboš pointed out, your sequence $(v_i)$ does not converge weakly to zero. The above criterion is not applicable, as your sequence is not bounded. Indeed, it's the canonical example showing that assuming boundedness is indeed necessary in that criterion.

Since you said that the uniform boundedness principle is still a bit of a mystery to you, I can't do better than recommend Alan Sokal's recent article A really simple elementary proof of the uniform boundedness theorem in which he gives a proof that gets away without using any Baire-trickery.

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Thank you! The professor probably did mention the necessity of $u_i$ being bounded, but as the whole counterexample was more of a sidenote anyway I missed it. – leftaroundabout May 19 '11 at 12:58
@leftaroundabout: You're welcome. Do have a look at Sokal's paper, it is really cool! – t.b. May 19 '11 at 13:01

Dear leftaroundabout, your $v_i$ sequence is not weakly convergent to zero. Take its inner product with $$x = (1/1, 1/2, 1/3, 1/4, 1/5, \dots )$$ to see that this inner product of $x$ with $v_i$ is actually equal to one in the infinite $i$ limit (and not zero as required by the weak convergence to zero). Note that $x$ is $l^2$ summable, because $\sum 1/n^2 = \pi^2/6$ is convergent, i.e. that it is a vector in the Hilbert space.

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Right... but that means it's actually not sufficient to show convergence of $\langle \varphi_j,v_i\rangle$ for all $\varphi_j$. How can I then proove any not-strongly convergent sequence at all to be weakly convergent? – leftaroundabout May 19 '11 at 10:59
If I understand you well: on the contrary: for the weak convergence it is enough to show the convergence of the inner product with any $\phi_i$ because that's how the weak convergence is defined. What I tried to argue is that you did not prove the convergence of the inner products to the "right" inner product and you could not because this convergence of inner products is not true. – Luboš Motl May 19 '11 at 11:07
This last comment doesn't make any sense to me. What is this "right" inner product you're talking about? – t.b. May 19 '11 at 11:12
@Theo: I believe he is referring to the inner product with $0$. That is, if the sequence converges weakly to zero, then the sequence of inner products with a vector $v$ should converge to $\langle 0,v\rangle=0$, so $0$ would be the "right" inner product to which the sequence of inner products should converge, and this is not the case for the example given above. (I know that the mathematics is clear to you, but hopefully that clears up Luboš's meaning for you. Apologies, Luboš, if I have misinterpreted.) – Jonas Meyer May 19 '11 at 11:30
@Jonas: Ah, okay, that is one viable way to make sense out of that comment. Thanks! I was seriously puzzled. – t.b. May 19 '11 at 11:38

The argument you attribute to your professor is incorrect.

Suppose we wish to show a sequence $(f_i)$ of vectors in $\ell^2$ (or any Hilbert space) is weakly convergent to some $f$. It is not sufficient to show that $$\langle f_i, \phi \rangle \to \langle f , \phi \rangle \quad \text{ as } i \to \infty \quad (*)$$ for all $\phi$ in some complete orthonormal set. Indeed, your counterexample $v_i$ shows that this is not sufficient. Nor is it sufficient to show that (*) holds for all $\phi$ in some dense set.

However, if one knows a priori that the set $\{f_i\}$ is bounded in $\ell^2$ (i.e. there exists $M$ with $||f_i|| \le M$ for all $i$), then it is sufficient to show that (*) holds for all $\phi$ in some dense set, or even for all $\phi$ in some set whose linear span is dense. Such as, for instance, a complete orthonormal set. Proving this is a good exercise.

It can be shown, using an appropriate version of the uniform boundedness principle, that even without knowing $\{f_i\}$ is bounded, it is sufficient to verify that (*) holds for all $\phi$ in some nonmeager set. However, nonmeager sets are not so easy to come by. (In particular, the linear span of a countable set is always meager.)

So to give a better proof that the sequence $\{u_i\}$ converges weakly (to 0), one could first note that the sequence is bounded in $\ell^2$ norm, and then check that (*) holds for all the $\varphi_j$ you describe (and aren't the $u_i$ and $\varphi_j$ actually the same here?). Actually, my preferred proof is to note that $u_i$ is itself an orthonormal set, and so for any $\phi \in \ell^2$ we have $$\sum_i |\langle u_i , \phi \rangle|^2 \le ||\phi||^2 < \infty$$ by Bessel's inequality. Since the sum converges we must have $\langle u_i , \phi \rangle \to 0$.

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Thank you! I guess it's not likely that I will ever encounter a sequence which is weakly convergent, but cannot easily be shown to be bounded? Because these nonmeager sets seem indeed rather scary to me. – leftaroundabout May 19 '11 at 13:12
@leftaroundabout: Of course, you can always try to show directly that $\langle f_i, \phi \rangle \to \langle f, \phi \rangle$ for all $\phi \in \ell^2$. My point about nonmeager sets is that although it is technically possible to verify it on a smaller set, it may not be practical, so it is not worth spending a lot of time trying to find a "good" smaller set to use. – Nate Eldredge May 19 '11 at 14:55
I've since learned that the comment about nonmeager sets is really not very useful at all. The set of $\phi$ for which (*) holds is obviously a vector space, hence showing (*) holds for all $\phi$ in some set $E$ is tantamount to showing it for the span of $E$. So we might as well assume $E$ is a vector space. But nonmeager proper subspaces of a Hilbert space are very weird. One can show that such a space cannot have the Baire property; in particular it is neither Borel, analytic, nor coanalytic. – Nate Eldredge Aug 9 '12 at 3:01
See here for a proof that $E$ does not have the Baire property. We would have to use the axiom of choice in an essential way to even show that sets without the BP exist (it is consistent with ZF+DC that they do not). – Nate Eldredge Aug 9 '12 at 3:05