Let $\{e_n\}$ be an orthonormal basis in a Hilbert space $H$ and let $\{\lambda_n\}$ be a sequence of numbers.

Define the operator $$T:H \to H$$ by $$Tu=\sum^\infty_{n=1} \lambda_n \langle u,e_n \rangle e_n$$ where $u \in H$

I am trying to show that if $\lim_{n \to \infty} \lambda_n=0$ then $T$ is a compact operator.

Why does showing $$||Tu-T_k u||^2 \leq \sup_{n >k} \{|\lambda_n|^2\} ||u||^2$$ tell us that $|T-T_k u|| \to 0$?

Where $T_k$ is $k^{th}$ partial sum in the definition of $T$.

I have the full proof for this but I am wondering why does showing $\lim_{n \to \infty} \lambda_n=0$ mean that $T$ is a compact operator?

How do you show that a operator is compact?

I have come across that statement that a compact operator is the limit of finite dimensional operators but I do not understand this statement.

What does this statement mean?

  • 1
    $\begingroup$ Hint: In a Hilbert space a compact operator is the (norm) limit of finite dimensional operators. $\endgroup$ – Nigel Overmars May 2 '16 at 21:36
  • $\begingroup$ @nigelovermars what does this statement mean, I dont understand $\endgroup$ – Al jabra May 2 '16 at 21:42

Let $X,Y$ be Hilbert spaces and let $T:X \to Y$ be a bounded linear operator, then we say that $T$ is compact if one, and hence all, of the following conditions hold:

  1. $T$ is the norm limit of a sequence $(T_n)$ of finite rank operators;
  2. For every bounded subset $V$ of $X$, $TV$ is relatively compact in $Y$;
  3. For each bounded sequence $(x_n)$ in $X$, there is a subsequence of $(Tx_n)$ which converges in $Y$.

In example, the first criterion is used: Let $T:H \to H$ be defined by $$Tu = \sum_{n=1}^\infty\lambda_n(u,e_n)e_n, \quad u \in H$$ and define $T_k :H \to H$ by $$T_ku = \sum_{n=1}^k\lambda_n (u,e_n)e_n$$ We will show that $$\|T - T_k\| \to 0, \quad n \to \infty$$ To do so, let $u \in H$, then you already have that $$\|Tu - T_ku\|^2 \leq \sup_{n \gt k} \{ |\lambda_n|^2\}\|u\|^2$$ which says, by definition, that $$\|T-T_k\|^2 \leq \sup_{n \gt k} \{ |\lambda_n|^2\}$$ Since $\lambda_n \to 0$, we have that $$\sup_{n \gt k} \{ |\lambda_n|^2\} \to 0, \quad k \to \infty$$ In other words, $$\|T-T_k\|^2 \to 0, \quad k \to \infty$$ And hence we conclude that $T$ is compact.

  • $\begingroup$ I am forgetting some real analysis here ,how do we know that $\sup_{n >k} \{|\lambda_n|^2\} \to 0$? $\endgroup$ – Al jabra May 2 '16 at 22:18
  • $\begingroup$ Also do we have a finite rank operator? $\endgroup$ – Al jabra May 2 '16 at 22:18
  • $\begingroup$ @Aljabra For finite rankness, observe that $\{e_1,...,e_k\}$ is a basis of $\operatorname{ran} T_k$. For your other question, by definition $\lambda_n \to 0$ implies that for all $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that $\lambda_n \in (-\varepsilon, \varepsilon)$ for all $n \geq N$. The $\sup$ of this set is $\varepsilon$, hence the $\sup$ of the limit is $0$. $\endgroup$ – Nigel Overmars May 3 '16 at 15:43
  • $\begingroup$ @NigelOvermars It is my understanding that $T$ is compact iff $lim(\lambda_n) \rightarrow 0$, but I can't seem to prove it. I would imagine it involves supposing that $lim(\lambda_n$ does not go to $0$, in which case $lim(\lambda_n) \rightarrow c \in \mathbb{R}, c \neq 0$, but I am at a loss for how to proceed $\endgroup$ – Vinny Chase May 9 '18 at 22:34

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