Schur's test : an infinite matrix defines a bounded operator on $\ell^2$

Question

I want to prove Schur test. Schur test says

Let $\{\alpha_{ij}\}_{i,j=1}^{\infty}$ be an infinite matrix such that $\alpha_{ij} \ge 0$ for all $i,j$ and such that there are scalars $p_i>0$ and $\beta,\gamma>0$ with $\sum_{i=1}^{\infty}\alpha_{ij}p_{i} \le \beta p_j \quad$ , $\quad \sum_{j=1}^{\infty}\alpha_{ij}p_{j} \le \gamma p_i \quad$ for all $i,j\ge 1$ then there is an operator $A$ on $\ell^{2}(\mathbb{N})$ with $\langle Ae_{j},e_{i}\rangle =\alpha_{ij}$ and $\|A\|^2 \le \beta \gamma$.

I know that I must use the orthogonal projection and Fourier series, But I don't know how. This is an exercise from section1, Chapter2, A Course in Functional Analysis Conway(second edition).

Answer 1

Let $\{e_j\}$ be the canonical basis and write $x=\sum_j x_je_j$. Assume first that $x_j=0$ when $j\geq j_0$ for a certain $j_0$. Then (note that everything is positive in the sums at the end, so we may change the order of summation in the second to last line; and the exchange of sums in the second line is trivial because the sums in $j$ and $k$ are actually sums, and not series) \begin{align} \|Ax\|^2&=\langle Ax,Ax\rangle =\sum_{k,j} \overline{x_k}x_j\langle Ae_k,Ae_j\rangle =\sum_{k,j} \overline{x_k}x_j\sum_{s,t}\alpha_{sk}\alpha_{tj}\langle e_s,e_t\rangle\\ \ \\ &=\sum_{k,j} \overline{x_k}x_j\sum_{t}\alpha_{tk}\alpha_{tj} =\sum_t\left| \sum_{j}\alpha_{tj}x_j\right|^2 =\sum_t\left| \sum_{j} \alpha_{tj}^{1/2}p_j^{1/2}\,\alpha_{tj}^{1/2}\frac{x_j}{p_j^{1/2}}\right|^2\\ \ \\ &\leq \sum_t\left(\sum_{j}\alpha_{tj}p_j \right)\,\left(\sum_{j}\alpha_{tj}\frac{|x_j|^2}{p_j} \right)\\ \ \\ &\leq \sum_t \gamma p_t\sum_j\alpha_{tj}\frac{|x_j|^2}{p_j} =\sum_j\gamma\frac{|x_j|^2}{p_j}\sum_t\alpha_{tj}p_t\\ \ \\ &\leq \sum_j\gamma\frac{|x_j|^2}{p_j} \,\beta p_j =\beta\gamma\,\|x\|^2. \end{align} As the elements $x$ with finite support are dense, the inequality $\|Ax\|^2\leq\beta\gamma\|x\|^2$ extends to all $x$.