Condition on the kernel of the integral operator to belong to the trace class? Let $\mu$ be a finite compactly supported Borel measure on the real line. Consider the integral operator $K$ on $L^2(\mu)$,
$$
(Kh)(x)=\int h(y)k(x-y)\, d\mu(y),
$$
where $k$ is a fixed function.
Question. $\,$ Under what condition on $k$ does the operator $K$ belong to the trace class?
As far as I understand, it suffices that $k$ belong to some $C^\alpha$ with $\alpha>0$. I am interested in a reference to the sharpest known result.

My own version is as follows: if $k$ is the Fourier transform of a certain function from $L^1$, then $K$ belongs to the trace class.
Proof. $\,$ Suppose that $k=\mathcal F u$ with $u\in L^1$. Then 
$$
k(x-y)=\int_{-\infty}^{+\infty} u(t)\exp(it(x-y))\,dt
=\int_{-\infty}^{+\infty} u(t)\exp(itx)\exp(-ity)\,dt.
$$ 
For $t\in\mathbb R$, denote by $A_t$ the integral operator with kernel $\exp(it(x-y))$. Then $A_t$ is a rank-one operator: $A_t=(\cdot, \exp(itx))\exp(itx)$. Each $A_t$ has norm $\mu(\mathbb R)$, hence the norm of $K=\int_{-\infty}^{+\infty}A_t\,u(t)\,dt$ in the trace class is not greater than $\mu(\mathbb R)\cdot \|u\|_{L^1}$.
