Linear operators well-defined by matrices in $l^2$ are bounded Let $\{a_{jk}\}$ be  an infinite matrix such that corresponding mapping $$A:(x_i) \mapsto (\sum_{j=1}^\infty a_{ij}x_j)$$ is well defined linear operator $A:l^2\to l^2$.
I need  help with showing that this operator will be bounded. I guess it means that i need to check if a unit sphere maps to something bounded, so i need to manage to get some inequality on coefficients of matrix that will allow to write a straight sequence of inequalities and get desired bound. But I don't understand how to get bound from operator being well defined.
This follows from the fact that the pointwise limit of bounded operators is bounded, which follows from the uniform boundedness principle:
If a sequence of bounded operator converges pointwise, then it is bounded in norm
It is helpful to consider the 'finite rank' case first:
Lemma. Let $(a_n)$ be a sequence of real numbers such that for every $x = (x_n) \in \ell^2$ we have $Ax := \sum_{n = 1}^\infty a_n x_n$ convergent. Then $A : \ell^2 \to \mathbb R$ is a bounded linear map.
Proof: Suppose $A$ is not bounded. Take any sequence of positive real numbers $(M_n)$. We may find $x_1 \in \ell^2$ with $\Vert x_1 \Vert = 1$ and $Ax_1 &gt; M_1$. We may find a finite subsequence $x_1'$ of $x_1$ satisfying $Ax_1' \geq M_1/2$. It is no restriction to assume all $a_n x_{1, n} \geq 0$. Because $x_1'$ is a subsequence, $\Vert x_1' \Vert \leq \Vert x_1 \Vert \leq 1$. Let $A_1$ be the corresponding subsequence of $(a_n)$. Then $A_1$ is a finite matrix, so it defines a bounded operator.
Because $A - A_1$ is also unbounded, we may find $x_2 \in \ell^2$ with $\Vert x_2 \Vert = 1$ and $(A-A_1)x_2 &gt; M_2$. We may find a finite subsequence $x_2'$ of $x_2$ with $(A-A_1) x_2' &gt; M_2/2$ and all $a_n x_{2,n} \geq 0$, and a corresponding subsequence $A_2$ of $A - A_1$. By construction, we may assume that $x_1'$ and $x_2'$ have disjoint supports, so that $(A-A_1) x_2' = A x_2'$.
Continuing this procedure, we find a sequence $(x_n')$ with disjoint supports, $\Vert x_n \Vert \leq 1$ and $A x_n' &gt; M_n/2$ and $a_m x_{n,m} \geq 0$. By taking the $M_n$ to grow sufficiently fast, the element $x := \sum_{n=1}^\infty \frac1{M_n} x_n'$ is in $\ell^2$. But because the $x_n'$ have disjoint supports, and because $Ax$ is a sum of nonnegative terms, it is clear that
$$Ax \geq \sum_{n=1}^\infty \frac1{M_n} A x_n &gt; \sum_{n=1}^\infty \frac12 = \infty \,,$$
so that $A$ is not well-defined. $\square$
Corollary. Let $(a_{ij})$ be a matrix of real numbers with $N$ rows, such that for every $x = (x_j) \in \ell^2$ we have $Ax := \left(\sum_{j = 1}^\infty a_j x_j \right)_i \in \mathbb R^N$ convergent. Then $A : \ell^2 \to \mathbb R^N$ is a bounded linear map.
Proof: Each of the projections of $A$ is bounded by the previous lemma, and this suffices because $\mathbb R^N$ is finite-dimensional. $\square$
Now, for the case of $A : \ell^2 \to \ell^2$.
Proof: Suppose $A$ is not bounded. Take any sequence of positive real numbers $(M_n)$. We may find $x_1 \in \ell^2$ with $\Vert x_1 \Vert = 1$ and $\Vert A x_1 \Vert &gt; M_1$. There then exists a submatrix $A_1$ of $A$ with finitely many nonzero rows, and with $\Vert A_1 x_1 \Vert$ 'very close' to $\Vert A x_1 \Vert$.
By the corollary above, $A_1$ is bounded, so that $A - A_1$ is not bounded. Thus we may find $x_2 \in \ell^2$ with $\Vert x_2 \Vert = 1$ and $\Vert (A-A_1) x_2 \Vert &gt; M_2$, and subsequently a submatrix $A_2$ of $A - A_1$ with finitely many rows and with $\Vert A_2 x_2 \Vert$ 'very close' to $\Vert (A-A_1) x_2 \Vert$.
Continuing this process, we find a sequence $(x_n) \in \ell^2$, with $\Vert x_n \Vert = 1$ and $\Vert A_n x_n \Vert$ 'very close' to $\Vert (A - A_1 - A_2 - \ldots - A_{n-1}) x_n \Vert$ and with $\Vert (A - A_1 - A_2 - \ldots - A_{n-1}) x_n \Vert &gt; M_n$. Note how the $A_n x_n$ have disjoint supports, by construction.
Consider the element $x = \sum_{n=1}^\infty \frac1{M_n} x_n$. When the $M_n$ grow sufficiently fast, this lies in $\ell^2$. Using orthogonality of the $A_n x$ and the triangle inequality applied to finite sums,
$$\begin{align*}\Vert Ax \Vert
&amp;\geq \sum_{n=1}^\infty \Vert A_n x \Vert \\
&amp;\geq \sum_{n=1}^\infty \left( \frac1{M_n}\Vert A_n x_n \Vert - \left \Vert A_n \sum_{j \neq n} \frac1{M_j} x_j \right\Vert \right) \\
&amp;\geq \sum_{n=1}^\infty \left( \frac1{M_n}\Vert A_n x_n \Vert - \sum_{j &lt; n} \frac1{M_j} \Vert A_n x_j \Vert - \Vert A_n \Vert \left \Vert \sum_{j&gt; n} \frac1{M_j} x_j \right\Vert \right) \\
&amp;\geq \sum_{n=1}^\infty \left( \frac1{M_n}\Vert A_n x_n \Vert - \sum_{j &lt; n} \frac1{M_j} (\Vert (A - A_1 - A_2 - \ldots - A_{j-1})  x_j \Vert - \Vert A_j x_j \Vert) - \Vert A_n \Vert \left( \sum_{j &gt; n} \frac1{M_j^2}\right)^{1/2} \right) \,.
\end{align*}$$
Now we choose the $M_n$ and $A_n$ so that this sum is infinite: At every step, take $M_{n}$ very large compared to the norms $\Vert A_j \Vert$ for $j &lt; n$. Given this $M_n$, choose $A_n$ so that $\Vert A_n x_n\Vert$ is extremely close to $\Vert (A - A_1 - A_2 - \ldots - A_{n-1}) x_n \Vert$. It is then clear that the negative terms in the above sum can be made arbitrarily small, to get (for example)
$$\Vert Ax \Vert
\geq \sum_{n=1}^\infty \frac1{M_n}\Vert A_n x_n \Vert/2 \geq \sum_{n=1}^\infty \frac12  = \infty \,.$$
More concretely, you could take $M_1 = 1$, $M_n&gt; \max(n, 10n \max_{j &lt; n}(\Vert A_j \Vert))$ and $A_n$ such that $\Vert (A - A_1 - A_2 - \ldots - A_{n-1}) x_n\Vert - \Vert A_n x_n \Vert \leq \frac1{100n^2}$. Then
$$\begin{align*}
&amp;\sum_{n=1}^\infty \left( \frac1{M_n}\Vert A_n x_n \Vert - \sum_{j &lt; n} \frac1{M_j} (\Vert (A - A_1 - A_2 - \ldots - A_{j-1}) x_j \Vert - \Vert A_j x_j \Vert) - \Vert A_n \Vert \left( \sum_{j &gt; n} \frac1{M_j^2}\right)^{1/2} \right) \\
&amp; \geq \sum_{n=1}^\infty \left( \frac9{10} - \sum_{j &lt; n} \frac1{100j^2} - \left( \sum_{j &gt; n} \frac1{100n^2}\right)^{1/2} \right)
\end{align*}$$
and this is fine.