I am studying functional analysis and in particular Hilbert spaces and operators, I would like some clarification on the idea of compact operators.
The definition that was presented to me is:
Given $\mathcal{H}_1$ and $\mathcal{H}_2$ Hilbert spaces on the field $\mathbb{K}$, an operator $T\in \mathcal{L}(\mathcal{H}_1,\mathcal{H}_2)$ is said to be compact if the immage of the ball $\overline{B_1}$ in $\mathcal{H}_1$ has compact closure in $\mathcal{H}_2$.
Also, how would one show that an equivalent definition is that, in the same environment, $T$ is compact if and only if: for all bounded sequences $\{x_n \}_{n\in \Bbb N}$ in $\mathcal{H}_1$, then the sequence of the immages $\{ Tx_n\}_{n\in \Bbb N}$ has a convergent subsequence in $\mathcal{H}_2$?
Thanks in advance.
If the image of the ball has compact closure, then the image of any bounded set has a compact closure as well (try to understand why). But in a metric space, being precompact (having compact closure) means that any sequence has a subsequence which converges (maybe to a point in the closure).
For the other direction - if the image of any bounded sequence has a convergent subsequence, then this is in particular true for the ball. Then the image of the ball is sequentially precompact, which in a metric space, is equivalent to being precompact (so the closure is compact).
Bottom line - since any Hilbert space is in particular a metric space, being (pre)compact and sequentially (pre)compact is the same, and there is no real difference between the ball and any other bounded set.