Showing a linear operator is compact

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Homework Statement



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

Then show that [itex]Ax=\sum_{j=1}^{\infty} \lambda_{j}(x,e_{j})e_{j}[/itex]

Defines a compact self adjoint operator [itex]H \rightarrow H[/itex]


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 [itex]x_{n}[/itex] and show that [itex]Ax_{n}=\sum_{j=1}^{\infty}\lambda_{j}(x_{n},e_{j})e_{j}[/itex] has some subsequence [itex]Ax_{n_{i}}[/itex] 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 [itex]x_n[/itex] to some subset ([itex]Ax_{n}>0[/itex], or if that is empty, [itex]Ax_{n} \leq 0[/itex]) and using that [itex]\lambda_{k} \rightarrow 0[/itex] 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.
 

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