Here is the link to the solution:

Why should we use the uniform boundedness principle here?

And here is the solution:

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][1].

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 $\|A\| \leq \sup_N \|T_N\|$.

But I do not understand this step:

$\sum_{i=1}^N \left|\sum_j a_{ij}x_j\right|^2 \leq \sum_{i=1}^N\left(\sum_j |a_{ij}|^2 \right)\left(\sum_j |x_j|^2 \right)$

I know that this is by Cauchy-Schwartz, but should not $a_{ij}x_j$ be inside an absolute value to apply Cauchy Schwartz?


It is the entire inner product on the LHS that lies inside the absolute value for Cauchy–Schwarz, not the individual products. The following, which corresponds to what you think is true, is not Cauchy–Schwarz: $$\langle|\mathbf u|,|\mathbf v|\rangle^2\le\langle\mathbf u,\mathbf u\rangle\langle\mathbf v,\mathbf v\rangle$$ This is the real Cauchy–Schwarz: $$|\langle\mathbf u,\mathbf v\rangle|^2\le\langle\mathbf u,\mathbf u\rangle\langle\mathbf v,\mathbf v\rangle$$

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  • $\begingroup$ I am unable to get your point of view …. could you please explain it in terms of the terms given in our problem? $\endgroup$ – Mathstupid Apr 8 at 6:16
  • $\begingroup$ @Mathstupid Here $\mathbf u$ is the vector of $a_{ij}$ for a fixed $i$, and $\mathbf v$ is the vector of $x_j$. $\endgroup$ – Parcly Taxel Apr 8 at 6:18

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