Given a bounded operator on a Hilbert space $A \in \mathcal{B}(H)$, I am familiar with the fact that the Hilbert-Schmidt norm of an operator is lesser or equal to the trace class norm, i.e. $\|A\|_{\mathcal{HS}} \leq \|A\|_{\mathcal{TC}}$.
While working with weighted $\ell_2$ spaces, I was faced with an apparent contradiction of this fact. Let me be clear : I have total confidence in the inequality $\|A\|_{\mathcal{HS}} \leq \|A\|_{\mathcal{TC}}$. I believe instead that I've made a logical or computational mistake somewhere. However, I'm unable to spot it.
Consider a weight function $w: \mathbb{N} \rightarrow (0,\infty)$ and let
$H$ be the Hilbert space $\ell_2(\mathbb{Z} / m \mathbb{Z})$ endowed with the weighted euclidean norm
$$ \|f\|_w :=~ \sqrt{\sum\limits_{k =1}^m w(k)^2 |f(k)|^2}.$$
Let's denote by $\langle \cdot, \cdot \rangle_w$ the associated inner product.
If $\{e_n\}_{n \in \mathbb{N}}$ is the standard orthonormal basis of $\ell_2(\mathbb{Z} / m \mathbb{Z})$, then $\{u_n\}_{n \in \mathbb{N}}$ is an orthonormal basis of $H$ where $u_n := \frac{e_n}{\|e_n\|_w} =\frac{e_n}{w(n)}$.
Now consider $A \in \mathcal{B}(H)$ a bounded operator on $H$.
- The Hilbert-Schmidt norm of $A$ is defined by $\sqrt{\sum\limits_{k=1}^m \big\|A(u_k)\big\|_w^2}$.
- The Trace Class norm of $A$ is defined by $\sum\limits_{k=1}^m \Big\langle (A^*A)^{\frac{1}{2}}(u_k),u_k\Big\rangle_w$, where $A^*$ is the adjoint operator of $A$ and $(A^*A)^{\frac{1}{2}}$ is a square root of the positive operator $A^*A$.
Now consider the operator $A:= \langle \cdot, e_i\rangle_w e_j$ for some $1 \leq i, j \leq m$.
I've computed that $\|A\|_{\mathcal{TC}} = w(i)^2$ whereas $\|A\|_{\mathcal{HS}} = w(i)w(j)$ but this is impossible because for $i \neq j$ we could find values of $w(i)$ and $w(j)$ for which $\|A\|_{\mathcal{HS}} > \|A\|_{\mathcal{TC}}$.
Hence there most be something wrong with my computations of the norms of the operator $A:= \langle \cdot, e_i\rangle_w e_j$. The thing is : when I look at my computations, I can't identity when I have sinned.
Trace Class norm : Observe first that $A$ can be identified with the $m$ by $m$ matrix whose entries are all zero except for the $(j,i)$-th which is $w(i)^2$. Hence $A^*A$ can be identified with the $m$ by $m$ matrix whose entries are all zero except for the $(i,i)$-th which is $w(i)^4$. Hence $(A^*A)^{\frac{1}{2}}$ can be identified with the $m$ by $m$ matrix whose entries are all zero except for the $(i,i)$-th which is $w(i)^2$. Therefore $$(A^*A)^{\frac{1}{2}}(u_k) = \begin{cases}w(i)e_i~~~ \text{if $k = i$}\\ 0~~~~~~~ \text{otherwise}
\end{cases}$$
Hence $$\|A\|_{\mathcal{TC}} = \sum\limits_{k=1}^m \Big\langle (T^*T)^{\frac{1}{2}} u_k, u_k \Big\rangle_w = \Big\langle (T^*T)^{\frac{1}{2}} u_i, u_i \Big\rangle_w = \big\langle w(i) e_i, u_i \big\rangle_w = \|e_i\|_w^2 = w(i)^2.$$Hilbert-Schmidt norm : $$\Big\|\langle \cdot, e_i\rangle_w e_j\Big\|_{\mathcal{HS}} = \left(\sum\limits_{k=1}^m \big\|\langle u_k, e_i\rangle_w
e_j\big\|_w^2\right)^{\frac{1}{2}} =\Big(w(i)^2
w(j)^2\Big)^{\frac{1}{2}} = w(i)w(j)$$ since $\langle u_k,
e_i\rangle_w = \left\langle \frac{e_k}{w(k)}, e_i\right\rangle =
\begin{cases}w(i)~~~ \text{if $k = i$}\\ 0~~~~~~~ \text{otherwise}
\end{cases}$.
Best Answer
Your matrix elements for the adjoint are not correct. You are seemingly calculating the matrix elements for $A$ in the basis $(e_i)_i$ which is not orthonormal. Using your scalar product it should then be $w_j^2$ for the adjoint in the $(i,j)$-th place. One has: $$ \langle A^* e_m , e_k \rangle = \langle e_m,A e_k\rangle = w_m^2 w_k^2 \delta_{m,j}\delta_{k,i}$$
The symmetry becomes perhaps even more apparent if you work in the basis $(u_i)_i$ of unit vectors for your norm. Write: $$ A = A_{i,j} = w_i w_j \langle \cdot, u_i \rangle u_j$$ and the adjoint is simply $A^* = A_{j,i}$. Thus $A^* A = w_i^2 w_j^2 \langle \cdot, u_i \rangle u_i$ from which you get that the two norms are identical. This is also as one should expect since $A$ has rank 1.