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Let $Lambda(x)=(lambda_1x_1,lambda_2x_n,...)$ be an operator $l_2 to l_2$. Show its range is closed iff $inf_{lambda_knot=0} |lambda_k|>0$

Mathematics Asked on December 20, 2021

Let $Lambda(x)=(lambda_1x_1,lambda_2x_n,…)$ be an operator $l_2 to l_2$. Show its range is closed iff $inf_{lambda_knot=0} |lambda_k|>0$.

I proved the backwards direction. If $x_n$ is Cauchy in $Range(Lambda)$ then $Lambda(y_n)=x_n$, $y_n$ is cauchy and so its limit will converge to the limit of $x_n$. I am having trouble proving the forward direction. I think it should be done by contradiciton. We assume the range is closed and yet there is a subsequence of $lambda_k$ cinverging to $0$. I was not succcesful in

One Answer

Hints: Let $M$ be the closed subspace generated by ${e_i: lambda_i neq 0}$ where $(e_i)$ is the standard orthonormal basis. (Note that $M$ consist of the sums $sum_{lambda_i neq 0} a_ie_i$ with $ sum |a_i|^{2} <infty$). Let $Lambda_1=Lambda |M$. Show that range of $Lambda_1$ is also closed and that $Lambda_1$ is also injective. By open mapping theorem its inverse is continuous so there exist a constant $C$ such that $sum_{lambda_i neq 0} |lambda_i x_i|^{2} geq C sum |x_i|^{2}$ for all $(x_i)$. From this the conclusion follows immediately.

Answered by Kavi Rama Murthy on December 20, 2021

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