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Here is the question:

Let $A = [a_{ij}]_{i,j = 1}^{\infty}$ be an infinite matrix of real numbers and suppose that, for any $x \in \ell^2,$ the sequence $Ax$ belongs to $\ell^2.$ Prove that the operator $T,$ defined by $T(x) = Ax,$ is a bounded operator on $\ell^2.$

My question is:

I got a hint to use the uniform boundedness principle here but I do not know why, could anyone explain this to me, please? what makes me when I look at a problem to decide that it should be solved by UBP?

EDIT:

1-I have taken this proposition: "The series $\sum_{n =1}^{\infty} a_{n} b_{n}$ converges absolutely for every convergent sequence $\{b_{n}\}$ iff $\sum_{n =1}^{\infty} |a_{n}|$ converges." will it be helpful here in our case? the problem is that here in our case we are in $l^2.$

2-Also, Should it be better to use the uniform boundedness principle or the following theorem to solve the problem given above?

Theorem:

Let $X,Y$ be Banach spaces and let $\{T_{n}\}_{n=1}^{\infty}$ and $T$ be operators in $\mathcal{L}(X,Y).$ then $\lim_{n} T_{n}x = Tx,$ for all $x \in X,$ iff

(a)the sequence $\{T_{n}\}$ is bounded;

(b)lim_{n} T_{n}x exists on a dense subset of $X.$

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    Let $A_n$ be the top-left $n\times n$ matrix (zeroing out everything else), then for each $x$, $\sup_n |A_n x|_2 < \infty$ and by the uniform boundedness principle, $\sup_n |A_n| < \infty$. Finally note that $Ax$ is the pointwise limit of $A_nx$. – user58955 Apr 06 '20 at 03:09
  • @user58955 could you provide more details, please? Also, what about my question above? How do I know that I should apply UBP? –  Apr 06 '20 at 03:15
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    Such questions are usually hard to answer. There is really no protocol. So, you want to show that $|Ax|$ is uniformly upper bounded for all unit vectors $x$. Now you are dealing with infinite sums and you need to use the fact that $Ax\in \ell^2$ for all $x$. Thus you may need to use special $x$'s to get something. It is a typical idea to approximate $A$ by finite-rank matrices (if you are taking a functional analysis course, you will see more examples like this when you're studying compact operators)... – user58955 Apr 06 '20 at 03:18
  • and here the easiest thing to do is to take $n\times n$ submatrix, corresponding to taking $x$ with only finitely many nonzero coordinates. Then you will see that UBP helps. – user58955 Apr 06 '20 at 03:19
  • why I should use special $x'$s and to get what? @user58955 –  Apr 06 '20 at 03:38
  • This is what is ad hoc. You plug in special values of $x$ to obtain something for the purpose of the problem. Here, since you have a $\ell_2$ sequence, it is again a typical attempt to approximate $x$ (which may consist of infinitely many nonzero coordinates) by its finite truncations: $(x_1,0,0,0,\dots), (x_1,x_2,0,0,\dots), (x_1,x_2,x_3,0,\dots)$. It may work and it may not work. But you can try that and see what you get afterwards. – user58955 Apr 06 '20 at 03:43
  • I got a hint to define $$T_{N}x = (\sum_{k=1}^{N}a_{ij}x_{j})$$ is this the same idea you are trying to convey to me? @user58955 –  Apr 06 '20 at 03:53
  • Yes, it's exactly the same as what I said. – user58955 Apr 06 '20 at 03:54
  • where is the truncation here? I can not see it. where is taking $x$ with finitely many nonzero terms?I can not see this in this hint.@user58955 –  Apr 06 '20 at 03:56
  • How can I verify that this $T_N$ is bounded?@user58955 –  Apr 06 '20 at 03:57
  • Oh, there is a slight difference between $T_N x$ and I said at the beginning. I were doing $n\times n$ truncation of the infinite matrix and the hint says it's $\infty\times n$. Anyway let's stick to your hint. Originally with infinitely many terms, it should be $\sum_{k=1}^\infty a_{ij}x_j$ and if you truncate $x$ to the first $N$ terms, the sum becomes $\sum_{k=1}^N a_{ij}x_j$ as in your hint – user58955 Apr 06 '20 at 08:25
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    You can use Cauchy-Schwarz to show that $T_N$ is bounded. – user58955 Apr 06 '20 at 08:26
  • I have taken this proposition: "The series $\sum_{n =1}^{\infty} a_{n} b_{n}$ converges absolutely for every convergent sequence ${b_{n}}$ iff $\sum_{n =1}^{\infty} |a_{n}|$ converges." will it be helpful here in our case? the problem is that here in our case we are in $l^2.$@user58955 –  Apr 06 '20 at 11:17

1 Answers1

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First let's show a simpler version (1-dimensional): If $\sum_i a_i x_i < \infty$ all for $x\in\ell^2$, then $a\in \ell^2$.

You can prove this claim using uniform boundedness principle, or you can just use Riesz Representation Theorem. See this post.

Now, let's go back to your problem. It follows from the claim above that each row of $A$ is in $\ell_2$. Define $T_N$ to be the restriction of $A$ onto the first $N$ rows, that is, $$ T_N x = \left(\sum_j a_{1j}x_j,\sum_j a_{2j}x_j,\dots,\sum_j a_{Nj}x_j,0,0,\dots,\right). $$ We claim that $\|T_N\| < \infty$. Note that $$ \|T_Nx\|_2^2 = \sum_{i=1}^N \left|\sum_j a_{ij}x_j\right|^2 \leq \sum_{i=1}^N\left(\sum_j |a_{ij}|^2 \right)\left(\sum_j |x_j|^2 \right) \leq \|x\|_2^2\cdot \sum_{i=1}^N\sum_{j=1}^\infty |a_{ij}|^2, $$ thus $$ \|T_N\| \leq \left(\sum_{i=1}^N\sum_{j=1}^\infty |a_{ij}|^2\right)^{1/2}. $$ (Note that the infinite sum over $j$ is finite because of the claim at the beginning.)

Now, for each fixed $x$, observe that $\|T_Nx\|_2$ is uniformly bounded by $\|Ax\|_2$ (since $\|T_Nx\|_2$ is just part of the sum for $\|Ax\|_2<\infty$). It follows from the uniform boundedness principle that $\sup_N \|T_N\|<\infty$. Note that $\|Ax\|_2 = \lim_{N\to\infty} \|T_Nx\|_2 \leq (\sup_N \|T_N\|)\|x\|$, which implies that $A$ is bounded and $\|A\| \leq \sup_N \|T_N\|$.

user58955
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  • In this sentence "(Note that the infinite sum over $j$ is finite because of the claim at the beginning.)" do you mean over $j$ or over $i$? –  Apr 07 '20 at 08:20
  • I mean over $j$. It's $\sum_{j=1}^\infty$, this is an infinite sum – user58955 Apr 07 '20 at 08:23
  • I think in the 8th line, in your middle term , the $i$ should be until $N.$ –  Apr 07 '20 at 08:23
  • Oh, you're right. Let me fix the typo – user58955 Apr 07 '20 at 08:24
  • which claim makes the infinite sum over $j$ finite ?I do not understand this. –  Apr 07 '20 at 08:28
  • I understood now, is this upon taking $x_{i} = 1$ for all $i$? –  Apr 07 '20 at 08:38
  • and because we have $Ax \in l^2$ for all $x \in l^2$ by assumption ..... am I correct? –  Apr 07 '20 at 08:41
  • I also have the same question about the infinite sum over $j$ being finite … could you clarify this please? –  Apr 07 '20 at 08:50
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    @Happy So it's slightly different from your hint. I am taking $T_N$ to be the first $n$ rows of $A$ while in your hint it is the first $n$ columns of $A$.

    No, I am not taking a special $x$ to say $\sum_j |a_{ij}|^2$ is finite. This is from the 1-dimensional version stated at the beginning of the proof. Since $Ax\in \ell^2$, every coordinate of $Ax$ must be finite. The $i$-th coordinate of $Ax$ is $\sum_j a_{ij}x_j$, and this is finite for any $x\in \ell^2$. The 1-dimensional version implies that the $i$-th row of $A$ is in $\ell^2$, that is, $\sum_j |a_{ij}|^2 < \infty$.

     – user58955 Apr 07 '20 at 09:16