I'm asking this purely out of curiosity. Any sequence of linear operators I have seen thus far has a very obvious limit and proving the convergence has not been much of a challenge either. So I'm looking for sequences of linear operators in a normed linear space, whose convergence to their respective pointwise limits is not easy to show. Please share with me any such sequence you have encountered.
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What about a sequence of operators in a Hilbert space? – Botond Feb 27 '20 at 17:17
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@Botond Yes of course that would be interesting as well – Smooth Alpert Frame Feb 27 '20 at 17:18
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This survey of Michael Lacey discusses the theorem of Carleson, a very difficult one in harmonic analysis, concerning the pointwise convergence of Fourier series:
https://arxiv.org/abs/math/0307008
This question&answers here on Math.SE is a kind of "baby Carleson theorem".
Giuseppe Negro
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In the linked Math.SE it is $$T_nf(x):= f(x+\tfrac1n).$$ In the Carleson one, it is more involved as it is essentially the Fourier transform. – Giuseppe Negro Feb 28 '20 at 15:18
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aren't you trying to prove the existence of a function f for which the sequence does not converge a.e. to f? – Smooth Alpert Frame Feb 29 '20 at 00:04
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I must thank you for introducing me to such an interesting example and all the related concepts I found subsequently. However, I was looking for a sequence of operators that does converge to a limit. – Smooth Alpert Frame Feb 29 '20 at 00:55
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1OK, these examples are perhaps too involved. Still, the Carleson theorem is about a sequence of operators that converge to a limit. – Giuseppe Negro Feb 29 '20 at 01:16
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Let $H$ be a real or complex Hilbert space and $T \in B(H)$ with $\lVert T \rVert \leqslant 1$, and let's define $A_n$ as $$A_nx:=\frac{1}{n+1}\sum_{k=0}^n T^k x \quad(\forall x \in H)$$ What's the pointwise limit of $A_n$?
(This is one of the extra exercises from the functional analysis course I've had, but the pointwise limit was given for us)
Botond
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1In this case, usually they say "limit in the strong operator topology", more than "pointwise limit", but anyway, that's just terminology. Nice exercise. I bet it converges to the projector on the space of fixed points of $T$. – Giuseppe Negro Feb 27 '20 at 17:51
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@GiuseppeNegro you can prove it if you have time and mood for it, especially as I didn't solve it myself :D – Botond Feb 27 '20 at 21:38
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1Well, I cannot go into details, but the idea is that you can decompose $x$ into three components; $$x=x_1+x_1' + x_r, $$where $Tx_1=x_1$, and $Tx_1'=e^{i\theta} x_1'$ , while $\lVert Tx_r\rVert\le \rho \lVert x\rVert$ for some $\rho<1$. So, $x_1$ is a $1$-eigenvector, $x_1'$ is an eigenvector corresponding to an eigenvalue of unit modulus, and $x_r$ is a remainder, corresponding to eigenvalues of modulus strictly smaller than $1$. This last piece will obviously converge to $0$, while $A_n x_1=x_1$. Thus, it remains to prove that $A_n x_1'\to 0$. This is used via the geometric series formula. – Giuseppe Negro Feb 28 '20 at 11:30
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@GiuseppeNegro If $T$ has only finitely many eigenvalues with $1$ absolute-value, then $\bigcup_n \text{ker}(T-\lambda_n I)$ is closed (where $|\lambda_n|=1$), so Riesz's decomposition theorem will guarantee that $x_r \perp x-x_r$, so from Pythagoras' theorem, we have that $||x||^2=||x_r||^2+||x-x_r||^2$, i.e. $||x_r|| \leqslant ||x||$, so $||Tx_r|| \leqslant ||x||$. But how can you get there the $\rho$, and what if there are infinitely many eigenvalues of $T$ with 1 absolute value? – Botond Mar 05 '20 at 11:54
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Yeah, I don't really know. The point is that, if $Tx=e^{i\theta}x$, with $\theta\ne 0$, then $A_n x=\frac{x}{n+1}\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\to 0.$ While if $Tx=x$ then $A_nx=x$ and if $\lVert Tx\rVert<\lVert x\rVert$ then $A_nx\to 0$. I bet these three cases are enough to exhaust the general case, but I don't really know the details right now. – Giuseppe Negro Mar 05 '20 at 13:48