If $H$ is an infinite dimensional (not necessarily separable) Hilbert space with $\dim H=\infty$, does there always exist a positive (i.e. self-adjoint and spectrum contained in $\mathbb{R}_{\geq0}$) compact operator $T$ with infinite rank and $\|T\|\leq1$?
I was trying to construct one with an orthonormal basis $E$, but I ended up with the identity operator $I\colon H\to H$, which is not compact.
Best Answer
If $H$ is your infinite-dimensional finite space, let $(e_n)_{n=1}^\infty$ be an orthonormal sequence (it does not need to be a basis, just norm-one vectors which are pairwise orthogonal).
The operator
$$T:H \to H \quad , \quad T(x) = \sum_{n=1}^\infty \lambda_n \langle x|e_n\rangle e_n$$
where $(\lambda_n)$ is a sequence of real numbers converging to $ 0$, is compact and self-adjoint. Its eigenvalues are $(\lambda_n)$ and $0$, in case $(e_n)$ is not a basis. Now for positivity, just pick $\lambda_n=\frac1n$, for example.
For computing the norm of $T$ you can use Bessel's inequality:
$$\left\|\sum_n \lambda_n \langle x|e_n \rangle e_n\right\|^2 \leq \max |\lambda_n|^2 \cdot \sum_n \langle x|e_n \rangle^2 \leq \max|\lambda_n|^2 \cdot \|x\|^2 $$
so in this case it is $\|T\|=1$.