Here's the question:

To let $ A = (a_ {ij}) _ {i, j = 1} ^ { infty} $ be an infinite matrix of real numbers and assume that for each $ x in ell ^ 2, $ the sequence $ Ax $ heard $ ell ^ 2. $ Prove that the operator $ T, $ defined by $ T (x) = Ax, $ is a restricted operator to $ ell ^ 2. $

** Here is my exam: **

We will use the principle of the uniformly limited theorem as we have shown that $ ( ell ^ 2, | x | _ {2}) $ is a standardized space. and it is easy to show that it is complete by showing that each Cauchy sequence also converges $ | . | _ {2} $, then $ ( ell ^ 2, | x | _ {2}) $ is a Banach room and we gave it to everyone $ x in ell ^ 2, $ the sequence $ Ax $ heard $ ell ^ 2. $ Which can be formulated mathematically as follows:

We can define according to the circumstances $ A: ell ^ 2 rightarrow ell ^ 2 $ by $$ A x = A ( xi_ {1}, xi_ {2}, …) = ( beta_ {1}, beta_ {2}, …), $$

Where (a_ {ij}) begin {equation *}

Left(

begin {array} {ccc}

xi_ {1} \

xi_ {2} \

vdots

end {array} right)

end {equation *}

=

Left(

begin {array} {ccc}

beta _ {1} \

beta_ {2} \

vdots

end {array} Law)

end {equation *}; i.e. $ beta_ {i} = sum_ {j = 1} ^ { infty} a_ {ij} xi_ {j} $.

$ | Ax | ^ {2} = sum_ {i = 1} ^ { infty} | beta_ {i} | ^ 2 < infty $

Therefore $ sup { | A x | } < infty $ and therefore $ sup { | T | } = sup { | A | } < infty $ as required.

Is my solution correct?