I'm beginner of functional analysis and few days ago, I learned about compact operators. If $H$ is a Hilbert space, we know that the space of finite rank operators $F(H)$ is not closed in the space of continuous (or bounded) linear operators $L(H)$ in general, and its closure is same as the space of compact operators $K(H)$. I find a different proof of the direction $K(H)\subseteq \overline{F(H)}$ and I want to know whether this is right or wrong.
Let $T\in K(H)$ be a compact operator. Then we can decompose $T$ as $$ T = \frac{1}{2}(T+T^{*}) + \frac{1}{2}(T-T^{*})=:T_{1}+T_{2}$$
where $T_{1}$ is self-adjoint and $T_{2}$ is anti-self-adjoint (I mean, $T_{2}^{*}=-T_{2}$), which follows from $T^{**}=T$. By the spectral theorem, $T_{1}$ can be written as
$$
T_{1} = \sum_{n=1}^{\infty} \lambda_{n}P_{n}
$$
where $\lambda_{n}$'s are eigenvalues of $T_{1}$ satisfying $\lim_{n\to \infty} \lambda_{n}=0$, and $P_{n}$'s are orthogonal projections to the eigenspaces $E(\lambda_{n})$, which are all finite dimensional (so $P_{n}$'s are finite rank operators). Similarly, since $T_{2}^{*}=-T_{2}$, $(iT_{2})^{*} = -iT_{2}^{*}= iT_{2}$, $iT_{2}$ is a self-adjoint compact operator, so we can apply the spectral theorem again and we get
$$
T_{2} = -i(iT_{2}) = -i\sum_{n=1}^{\infty} \lambda_{n}'P_{n}'
$$
where $\lambda_{n}'s$ are eigenvalues of $T_{2}$ with $\lim_{n\to\infty} \lambda_{n}'=0$ and $P_{n}'$'s are orthogonal projections to the eigenspaces $E(\lambda_{n}')$, which are finite dimensional spaces. Hence $$ \lim_{N\to\infty} ||T- \sum_{n=1}^{N}(\lambda_{n}P_{n}-i\lambda_{n}'P_{n}')||=0$$ and we get the result.
Best Answer
Your argument is correct. If you've never done it before, you should prove that $$ \lim_{N\to\infty}\left\|\sum_{n>N}\lambda_nP_n\right\|=0. $$ Or, more explicitly, that $$ \left\|\sum_{n>N}\lambda_nP_n\right\|=\sup\{|\lambda_n|:\ n>N\}. $$