Let me start by saying that I do appreciate any insight on this. So also if you have a partial result, please share it as a comment or answer.


This is somewhat unrelated to what I normally do, so I may be missing something rather obvious here, but unlike for Hilbert-Schmidt norms, very little useful methods seem to be available to calculate the norm of Trace-class operators.


Let $f \in C\_c^{\infty}(\mathbb{R}\times\mathbb{R}),$ then we have an integral operator


$$Tg(s):=\int\_{\mathbb{R}^2} f(s+t\_1,t\_2) g(t\_1,t\_2)dt$$ where $T:L^2(\mathbb{R}^2) \rightarrow L^2(\mathbb{R}).$


My goal is to compute its trace-norm or get at least some useful bounds on it. For this, it may be good to have its adjoint at hand 


$$T^\*h(t\_1,t\_2):= \int\_{\mathbb{R}}\overline{f(s+t\_1,t\_2)} h(s)ds.$$


However, now it is still completely unobvious to me how to compute $\sqrt{T^\*T}$ which is needed in the standard definition of the trace norm(=nuclear norm). 


Possible ways to calculate the nuclear norm that I could imagine to be useful include


a.) Use $\left\lVert T \right\rVert\_{\text{nuclear}}= \sum\_{n} \left\lvert \langle \sqrt{T^\*T}e\_n,e\_n \rangle \right\rvert$ for an arbitrary orthonormal basis $(e\_n)$


b.) Use Hahn-Banach, i.e. $\left\lVert T \right\rVert\_{\text{nuclear}} = \sup\_{S \in L(L^2(\mathbb{R}),L^2(\mathbb{R}^2)); \left\lVert S \right\rVert=1 } \operatorname{tr} (T^\*S)$ or 


c.) Use that $\left\lVert T \right\rVert\_{\text{nuclear}}$ is the supremum of all $\sum\_{n} \lvert\left\langle e\_n,Tf\_n \right\rangle \rvert$ for ONS $(e\_n),(f\_n).$ (this may be useful to get bounds)


**Questions:** I am paticularly curious to find out whether:


**1.)** There are any theorems or tricks that apply to this operator which allow me to compute its nuclear norm.


**2.)** Just based on intuition I would assume that the nuclear norm would be something like 
$$ \int\_{\mathbb{R}^2} \left\lvert f(t\_1+t\_1,t\_2) \right\rvert dt\_1 dt\_2 = \frac{1}{2} \left\lVert f \right\rVert\_{L^1}.$$
Can we say if this is at least a correct lower/upper bound for the trace-norm?
Maybe it is possible to find an operator $S$ or orthonormal bases (in the definition of the nuclear norms above) so that we get this as a lower bound, at least?


**3.)** Are there any non-trivial upper/lower bounds available?


If you have any questions about this, please let me know.



Trace-norm of integral operator
This is typically not a trace class operator. The problem is that the kernel
$$
K(t,u) = \int \overline{f(s+t\_1,t\_2)}f(s+u\_1,u\_2)\, ds
$$
of $T^\*T$ depends on the first coordinates only through the difference $t\_1-u\_1$, so has no uniform decay in these directions. It follows that $\int\!\!\int |K|^2\, dt\,du =\infty$ (for typical $f$ at least), so $T^\*T$ is not Hilbert-Schmidt and thus $(T^\*T)^{1/2}\notin B\_4$ (let alone trace class).


In fact, I think $T$ is usually unbounded, for similar reasons (or we can think of $T$ as convolution by $f$ in $\mathbb R^2$, followed by restriction to $\mathbb R \times \{ 0\}$, and this final operation is not bounded).


