When are integral operators trace class? Define an integral operator $T$ on $L^2([0,1])$ by $$Tf(x) = \int_0^1 K(x, y) f(y) \, dy. $$ Such an operator is Hilbert-Schmidt when $K$ is in $L^2([0,1]\times [0,1])$. 
I heard that if $K$ is smooth, then $T$ is in fact trace-class. Why is this? 
When is such an operator trace-class? 
 A: There is a treatment of this result in the book 'Functional Analysis' by Lax.
Note that $ T^{*}T $ also has a smooth kernel,
\begin{equation*}
\tilde{K}(x,y)=\int \bar{K}(z,x) K(z,y)\, dz,
\end{equation*}
and that the min-max principle ensures that the non-increasing sequence of eigenvalues of $T^{*} T$, $\lambda_1 \geqslant \lambda_2 \geqslant \ldots$, are given by
\begin{equation*}
  \lambda_{n} = \min_{\substack{\mathcal V \subseteq \mathcal H \\ \dim \mathcal V = n-1}} \max_{x\in \mathcal V^{\perp}} \, \langle x, T^{*} T x\rangle.
\end{equation*}
Thus, if $\mathcal V _{n}$ denotes the subspace of polynomials of degree $\leqslant n-2$, then we have
\begin{equation*}
  \lambda_{n} \leqslant \max_{x \in \mathcal V_{n} ^{\perp}} \, \langle x, T^{*} T x\rangle = \max_{x \in \mathcal V_{n} ^{\perp}} \, \langle x, (T^{*} T - L_{p_{n-2}}) x\rangle \leqslant \lVert \tilde{K} - p_{n-2}\rVert_{\infty} \rightarrow 0,
\end{equation*}
where $L_{p_{n-2}}$ denotes an integral operator whose kernel
$$
p_{n-2}(x,y)=\sum_{j=0}^{n-2}\sum_{i=0}^{n-2}c_{ij}x^i y^j
$$
is a polynomial of degree at most $n-2$ in each separate variable, and the sequence $(p_{n})_{n\geqslant 0}$ is a sequence chosen by means of the Weierstrass Approximation Theorem such that $\lVert \tilde{K} - p_{n-2}\rVert_{\infty} \rightarrow 0$. We conclude that $T^{*} T$ is compact, which implies that $T$ is compact. In order to show that $T$ is trace class, we must refine the argument in such a way that
\begin{equation}
  \tag{1}
  \sum_{n=1}^{\infty}\sqrt{\lambda_{n}} < \infty,
\end{equation}
i.e. we need to pick polynomials which approximate $K$ sufficiently well.
First, a reminder concerning Bernstein polynomials. Given a continuous function $f:[0,1]\rightarrow \mathbb{R}$, introduce the Bernstein polynomial of degree $n$,
\begin{equation*}
  p_{n}(t)=\sum_{m=0}^{n} f\left(\frac{m}{n}\right) \binom{n}{m} t^{m}(1-t)^{n-m}.
\end{equation*}
Then we have $p_{n}\rightarrow f$ uniformly. In fact, if we pick iid Bernoulli distributed variables $X_{1}^{(t)},X_{2}^{(t)},\ldots$, with $X_{j}^{(t)}\sim \text{Bernoulli}(t)$ and define $S_{n}^{(t)}=\frac{1}{n}\sum_{m=1}^{n} X_{m}^{(t)}$ then we have
\begin{equation*}
  p_{n}(t)=E f\left(S_{n}^{(t)} \right),
\end{equation*}
since $nS_{n}^{(t)} \sim \text{Binomial}(n,t)$. Furthermore, since $f$ is uniformly continuous, given $\epsilon>0$, we may pick $\delta>0$ such that if $\lvert f(t) - f(x) \rvert > \epsilon / 2 $, then $\lvert t - x \rvert > \delta$. It now follows from Chebyshev's inequality that
\begin{equation*}
  P(\lvert f(S_{n}^{(t)}) - f(t) \rvert\geqslant \epsilon / 2 ) \leqslant P(\lvert S_{n}^{(t)} - t \rvert\geqslant \delta)  \leqslant \frac{t(1-t)}{\delta^{2}n}.
\end{equation*}
Thus, since $t(1-1)\leqslant 1/4$, we find if $n>\lVert f \rVert_{\infty}/ \epsilon \delta ^{2}$ that
\begin{equation*}
  \lvert p_{n}(t) - f(t) \rvert \leqslant E\lvert f(S_{n}^{(t)}) - f(t) \rvert \leqslant \epsilon/2 + \frac{\lVert f \rVert_{\infty}}{2\delta^{2}n}<\epsilon,
\end{equation*}
which concludes the proof that $p_{n}\rightarrow f$ uniformly.
We now extend the construction to two variables, so that it may be applied to obtain an approximation of $K$. For $n\in \mathbb{N}_{0}$, define
\begin{align*}
  p_{n}(t_{1},t_{2})&=E f(S_{1,n}^{(t_{1})},S_{2,n}^{(t_{2})}) \\
&= \sum_{m_1=0}^{n} \sum_{m_2=0}^{n} f\left(\frac{m_1}{n},\frac{m_2}{n}\right) \binom{n}{m_1} \binom{n}{m_2} t_1^{m_1}(1-t_1)^{n-m_1} t_2^{m_2}(1-t_2)^{n-m_2},
\end{align*}
where $S_{1,n}^{(t)}=\frac{1}{n}\sum_{m=1}^{n}X_{1,m}^{(t_{1})},S_{1,n}=\frac{1}{n}\sum_{m=1}^{n}X_{2,m}^{(t_{2})}$ and $X_{1,m}^{(t_{1})},X_{2,m}^{(t_{2})}$ are iid with $\text{Bernoulli}(t_{1}),\text{Bernoulli}(t_{2})$ distribution. As before, given $\epsilon>0$, we pick $\delta>0$ such that if $\lvert f(t) - f(x) \rvert > \epsilon / 2 $, then $\lvert t - x \rvert > \delta$. Putting $t=(t_{1},t_{2})$ and $S_{n}^{(t)} = (S_{n}^{(t_{1})},S_{n}^{(t_{2})})$, we find similarly to before that
\begin{align*}
  &P(\lvert f(S_{n}^{(t)}) - f(t) \rvert\geqslant \epsilon / 2 ) \leqslant P(\lvert S_{n}^{(t)} - t \rvert\geqslant \delta) \\
  &\qquad \leqslant P(\lvert S_{1,n}^{(t_{1})} - t_{1} \rvert\geqslant \delta/2) + P(\lvert S_{2,n}^{(t_{2})} - t_{2} \rvert\geqslant \delta/2) \\
  &\qquad \leqslant \frac{4 t_{1}(1-t_{1})}{\delta^{2}n} + \frac{4 t_{2}(1-t_{2})}{\delta^{2}n} \leqslant \frac{6}{\delta^{2}n}.
\end{align*}
Thus, we find if $n> 32 \lVert f \rVert_{\infty}/ \epsilon \delta ^{2}$ that
\begin{equation*}
  \lvert p_{n}(t) - f(t) \rvert \leqslant E\lvert f(S_{n}^{(t)}) - f(t) \rvert \leqslant \epsilon/2 + \frac{16 \lVert f \rVert_{\infty}}{\delta^{2}n}<\epsilon,
\end{equation*}
which concludes the proof that $p_{n}\rightarrow f$ uniformly in this case.
At this point we have an explicit construction of an approximating sequence of polynomials $(p_{n})_{n\geqslant 0}$ at hand. In order to obtain (1), we refine this choice of approximating sequence. Recall the fifth order Taylor formula with remainder, namely
\begin{align*}
  f(t)&=\sum_{\lvert\alpha\rvert\leqslant 5} \, \partial^{\alpha}\! f(x)(t-x)^{\alpha}/\alpha! \\
  &\qquad + 6 \sum_{\lvert\alpha\rvert = 6} \int_{0}^{1}(1-s)^{5}\,\partial^{\alpha}\! f(ts+(1-t)x)(t-x)^{\alpha}/\alpha! \,ds.
\end{align*}
This suggests that we consider the refined sequence
\begin{equation*}
  q_{n+5}(t) = \sum_{ \lvert \alpha \rvert \leqslant 5 } E \, \partial ^{\alpha}\!f (S_{n}^{(t)}) (t-S_{n}^{(t)}) ^{\alpha} /  \alpha!.
\end{equation*}
Note that $q_{n+5}$ is a polynomial of degree at most $n+5$ in $t_1$ or $t_2$. This time around, we find
\begin{equation*}
  \lvert q_{n+5}(t) - f(t) \rvert \leqslant   \lVert \sum_{\lvert\alpha\rvert = 6} \, \partial^{\alpha}\! f/\alpha! \rVert_{\infty} \,E\,\lvert S_{n}^{(t)} - t\rvert^{6}.
\end{equation*}
By Minkowski's inequality, we have
\begin{equation*}
  (E\,\lvert S_{n}^{(t)} - t\rvert^{6})^{1/6} \leqslant (E\,\lvert S_{1,n}^{(t_{1})} - t_{1}\rvert^{6})^{1/6} + (E\,\lvert S_{n}^{(t_{2})} - t_{2}\rvert^{6})^{1/6},
\end{equation*}
and one can show that
\begin{equation*}
E\,\lvert S_{1,n}^{(t_{1})} - t_{1}\rvert^{6} \leqslant \frac{C}{n^{3}}
\end{equation*}
for some constant $C>0$. This leads to the bound
\begin{equation*}
  E\,\lvert S_{n}^{(t)} - t\rvert^{6} \leqslant \frac{C'}{(n+7)^{3}}
\end{equation*}
for another constant $C'>0$. All in all, we find
\begin{equation*}
  \lvert q_{n}(t) - f(t) \rvert \leqslant C''/(n+2)^{3},
\end{equation*}
and therefore $\lambda_{n} \leqslant C''/n^{3}$, which suffices to ensure that (1) holds.
