Showing a linear operator is compact

[SP90]

Homework Statement

Let $[e_{j}:j\in N]$ be an orthonormal set in a Hilbert space H and $\lambda_{k} \in R$ with $\lambda_{k} \rightarrow 0$

Then show that $Ax=\sum_{j=1}^{\infty} \lambda_{j}(x,e_{j})e_{j}$

Defines a compact self adjoint operator $H \rightarrow H$

The Attempt at a Solution

I've proven that it's self adjoint by extending the orthonormal set to an orthonormal basis and taking an arbitrary y in H and taking the inner product of that y written as a summation using the basis with Ax and then comparing that with the inner product of x with Ay.

To show that it's compact I need to consider some bounded sequence $x_{n}$ and show that $Ax_{n}=\sum_{j=1}^{\infty}\lambda_{j}(x_{n},e_{j})e_{j}$ has some subsequence $Ax_{n_{i}}$ which is convergent.

I'm not sure what direction to take here. Perhaps use the definition that A is compact if the image of the unit ball under A is sequentially compact? Or perhaps taking some the restriction of the sequence $x_n$ to some subset ($Ax_{n}>0$, or if that is empty, $Ax_{n} \leq 0$) and using that $\lambda_{k} \rightarrow 0$ and x_{n} is bounded to show that this subsequence is cauchy and therefore convergent.

I'm not confident with proving that linear operators are compact in general.

 

[mighty2000]

Any help or guidance would be appreciated.First, let's recall the definition of a compact operator. A linear operator A: H -> H is compact if for any bounded sequence (x_n) in H, there exists a subsequence (x_ni) such that Ax_ni converges in H.

Now, let's consider a bounded sequence (x_n) in H. Since (x_n) is bounded, the sequence (Ax_n) is also bounded. This means that the image of the unit ball under A, denoted by B_A, is bounded in H.

Next, we want to show that B_A is sequentially compact, i.e. any sequence in B_A has a convergent subsequence. Let (y_n) be a sequence in B_A. This means that ||y_n|| \leq 1 for all n and hence, (y_n) is a bounded sequence in H. Since H is a Hilbert space, by the Riesz Representation Theorem, there exists a unique sequence (z_n) in H such that y_n = (z_n, e_j) for all n, where (e_j) is an orthonormal basis for H.

Now, let's consider the sequence (Az_n) in H. Since (Az_n) is a sequence in the image of A, it can be written as Az_n = \sum_{j=1}^{\infty} \lambda_j (z_n, e_j) e_j. But we know that \lambda_j \rightarrow 0 as j \rightarrow \infty. This means that for any \epsilon > 0, there exists N such that for all j > N, |\lambda_j| < \epsilon. Therefore, for all n and j > N, we have |\lambda_j (z_n, e_j)| < \epsilon ||z_n|| \leq \epsilon. This means that (Az_n) is a bounded sequence in H.

Since H is a Hilbert space, by the Bolzano-Weierstrass Theorem, (Az_n) has a convergent subsequence (Az_ni) in H. But then, by the definition of A, (Az_ni) = \sum_{j=1}^{\infty} \lambda_j (z_ni, e_j) e_j. This means that for all j, \lambda_j (z_ni, e_j