$ι:U→V$ is an embedding, $Q:=ιι^*$, $L∈(ℝ^d)$, $Φ∈\text{HS}(U,ℝ^d)$ $⇒$ $\text{tr}LΦ\sqrt Q(Φ\sqrt Q)^*$ doesn't depend on $ι$ Let$^1$


*

*$U$ and $V$ be separable $\mathbb R$-Hilbert spaces

*$\iota\in\operatorname{HS}(U,V)$ be a Hilbert-Schmidt embedding

*$Q:=\iota\iota^\ast$

*$u:\mathbb R^d\to\mathbb R$ be twice Fréchet differentiable with second Fréchet derivative denoted by ${\rm D}^2u:\mathbb R^d\to\mathfrak L(\mathbb R^d)$, i.e. $${\rm D}^2u(x)y=\nabla^2u(x)y\;\;\;\text{for all }x,y\in\mathbb R^d$$ where $\nabla^2u(x)$ denotes the Hessian of $u$ at $x\in\mathbb R^d$

*$x\in\mathbb R^d$ and $L:={\rm D}^2u(x)$

*$\Phi\in\operatorname{HS}(U,\mathbb R^d)$ and $B:=\Phi\sqrt Q$



How can we calculate $\operatorname{tr}LBB^\ast$ and show that it doesn't depend on $\iota$?

I've tried the following: Since $Q$ is nonnegative and symmetric with finite trace, there is an orthonormal basis $(f_n)_{n\in\mathbb N}$ of $V$ with $$Qf_n=\lambda_nf_n\;\;\;\text{for all }n\in\mathbb N\tag 1$$ by the Hilbert-Schmidt theorem. Now, we can show that $$e_n:=\sqrt Qf_n\;\;\;\text{for }n\in N:=\left\{n\in\mathbb N:\lambda_n>0\right\}$$ is an orthonormal basis of $U$. Thus,
\begin{equation}
\begin{split}
\operatorname{tr}LBB^\ast&\stackrel{\text{def}}=\sum_{n\in\mathbb N}Bf_n\cdot LBf_n\\
&\stackrel{\text{def}}=\sum_{n\in\mathbb N}\Phi\sqrt Qf_n\cdot L\Phi\sqrt Qf_n\\
&=\sum_{n\in\mathbb N}\sqrt Qf_n\cdot\Phi^\ast L\Phi\sqrt Qf_n\\
&=\sum_{n\in N}e_n\cdot\Phi^\ast L\Phi e_n\\
&\stackrel{\text{def}}=\operatorname{tr}\Phi^\ast L\Phi
\end{split}\tag 2
\end{equation}
So, $\operatorname{tr}LBB^\ast$ doesn't depend on $\iota$ and $$\operatorname{tr}LBB^\ast=\operatorname{tr}\Phi^\ast L\Phi=\operatorname{tr}L\Phi\Phi^\ast\stackrel{\text{def}}=\sum_{i=1}^dL\Phi\Phi^\ast x_i\cdot x_i\tag 3\;,$$ where $(x_1,\ldots,x_d)$ denotes the Standard basis of $\mathbb R^d$.

Did I made any mistake so far? If not, can we further simplify the right-hand side of $(3)$?


$^1$ Let $\mathfrak L(A,B)$ and $\operatorname{HS}(A,B)$ denote the space of bounded, linear operators and Hilbert-Schmidt Operators, respectively. Moreover, let $\mathfrak L(A):=\mathfrak L(A,A)$.
 A: I am not 100% sure of the exact definition of the object you are using, but I will make soame commets anyway.
1) I think there is typo. The sense of the question seems to require $\Phi\in\operatorname{HS}(V,\mathbb R^d)$, and not $\Phi\in\operatorname{HS}(U,\mathbb R^d)$ as you write.  
2) $x, u$ ar decorations; all it matters is that $L \in\mathfrak L(\mathbb R^d)$, and $L^*=L$
3) The statement "$e_n:=\sqrt Qf_n\;\;\;\text{for }n\in N:=\left\{n\in\mathbb N:\lambda_n>0\right\}$" looks suspicious: $e_n \in U$, and $\sqrt Qf_n \in V$. There should be a $\iota$ or $\iota^*$ somewhere to place left hand side and right hand side on the same space. The relation 
$$ e_n:=\lambda_n^{-\frac{1}{2}}\iota^*(f_n)\;\;\;\text{for }n\in N:=\left\{n\in\mathbb N:\lambda_n>0\right\}$$
seem to work with the rest of the statement, if my understanding of HS embedding is correct.
With this relation, $\iota(e_n) = \lambda_n^{-\frac{1}{2}}\iota\iota^*(f_n) = \lambda_n^{-\frac{1}{2}}Q(f_n) = \lambda_n^{\frac{1}{2}}f_n, \; \iota^* \iota(e_n) =\lambda_n e_n $ and
$$ \sum_n ||\iota(e_n)||_V ^2 = \sum_n\lambda_n .$$
4) Now lets go over (2) in your proof. 
\begin{equation}
\begin{split}
\operatorname{tr}LBB^\ast&=\sum_{n\in\mathbb N}\sqrt Qf_n\cdot\Phi^\ast L\Phi\sqrt Qf_n\\
&=\sum_{n\in\mathbb N} <\lambda_n^{\frac{1}{2}} f_n| \Phi^\ast L\Phi \lambda_n^{\frac{1}{2}} f_n>_V\\
&=\sum_{n\in N} <\iota(e_n) | \Phi^\ast L\Phi \iota(e_n)>_V \\
&=\sum_{n\in N} \lambda_n^{-1} <\iota \iota^* \iota(e_n) | \Phi^\ast L\Phi \iota(e_n)>_V \\
&=\sum_{n\in N} \lambda_n^{-1} < \iota(e_n) | \iota (\iota^* \Phi^\ast L\Phi \iota \; e_n) >_V \\
&=\sum_{n\in N} \lambda_n^{-1} <e_n | \iota^* \Phi^\ast L\Phi \iota e_n >_{HS} \\
\end{split}
\end{equation}
The trace of $\iota^* \Phi^\ast L\Phi \iota $ is:
\begin{equation}
\begin{split}
\operatorname{tr}\iota^* \Phi^\ast L\Phi \iota &=
\sum_{n\in N} \lambda_n^{-1} <e_n , \iota^* \Phi^\ast L\Phi \iota \; e_n >_{U} \\
\end{split}\tag{a}
\end{equation}
whereas the trace of $LBB^*$ is
\begin{equation}
\begin{split}
\operatorname{tr}LBB^\ast&=\sum_{n\in N} <\iota(e_n) | \Phi^\ast L\Phi \iota(e_n)>_V \\
&=\sum_{n\in N} <e_n | \iota^* \Phi^\ast L\Phi \iota(e_n)>_U \\
\end{split}\tag{b}
\end{equation}
(a) and (b) are different.
It is easy to trace the computations when all spaces are finite dimentional, keeping track of the 3 different dot-products at play (i.e.: the dot product in $U$, then one in $V$, and the HS product induced on $U$ by the HS embedding.
