# problem on $L^2$ (pointwise) convergence and Carleson's Theorem

I am having some problem on my arguments and I want to see where it fails. It deals with pointwise convergence in $$L^2[-\pi,\pi]$$: pointwise convergence is given by Carleson's Theorem (so it is a hard result!) so my following argument must be wrong.

First, pick your function $$f\in L^2[-\pi,\pi]$$, and consider $$B=\{g_k\}_{k\geq 0}$$ to be the family or orthogonal functions $$\frac{1}{\sqrt{2\pi}}, \frac{1}{\sqrt{\pi}}\cos(nx)$$ and the corresponding normalized sinus term. We know that $$B$$ is complete, namely if $$(f,g)=0$$ for all $$g\in B$$, then necessarily $$f$$ is equal to $$0$$ (almost everywhere).

We also know (by Riesz-Fischer theorem) that $$L^2$$ is a complete space, so for any sequence $$\{c_k\}_{k\geq 0}$$ with $$\sum_{k\geq 0} c_k^2< \infty$$, the function

$$\bar{f}=\sum_{k\geq 0}c_k g_k$$

is inside $$L^2$$, and working just a little we see that for each $$k$$, $$(\bar{f},g_k)$$ is precisely $$c_k$$.

So, now, let us take $$f$$ with Fourier coefficients $$c_k=(f,g_k)$$, which, by Bessel, satisfies that $$\sum_{k\geq 0} c_k^2 <\infty$$. With these coefficients we can define the corresponding function $$\bar{f}$$.

Then, here it comes the crutial argument where I get lost/probably doing a mistake: as for every $$k$$ we have that $$(f,g_k)=(\bar{f},g_k)=c_k$$, then $$(f-\bar{f},g_k)=0$$. As $$B$$ is a complete family, this means that necessarily $$f=\bar{f}$$ almost everywhere.

But this is telling us precisely that the Fourier series of $$f$$ convergences pointwise almost everywhere to $$f$$, which is the content of Carleson's Theorem.

So, what I am doing wrong?

The sum $$\sum_{k\geq 0}c_kg_k$$ is convergent in $$L^2$$. A priori we have no reason to believe its $$L^2$$ limit is also its pointwise limit. Indeed, there exist convergent sequences in $$L^2$$ which are not pointwise convergent at any point.

Carleson's theorem in essense says that if you have a sequence of this special form, given by some linear combination of orthonormal basis, is going to be pointwise convergent almost everywhere to its $$L^2$$ limit.

Here is an example of a sequence as mentioned above: for each $$n\in\mathbb N$$ and $$0\leq k let $$f_{n,k}=\chi_{[k/n,(k+1)/n]}$$. Then the sequence $$f_{1,1},f_{2,1},f_{2,2},f_{3,1},f_{3,2},f_{3,3},f_{4,1},\dots$$ converges to $$0$$ in $$L^2$$ (since the $$L^2$$ norm of $$f_{n,k}$$ is $$1/\sqrt{n}$$), but this sequence is not pointwise convergent anywhere.

• A new one on me. I guess I should have studied the counterexamples footnotes more closely. Do you think you could reference a norm-convergent but everywhere divergent sequence? May 19 at 21:59
• @FShrike I have now included an example in the answer. May 19 at 22:05
• Nice! Thanks again+1 May 19 at 22:06