For every $A\in \mathcal{L}(C^\infty(\mathbb T^n))$ exists $k_A\in C^\infty(\mathbb T^n\times \mathbb T^n)$ such that..

Question

does anyone know whether it is true that for every $A\in\mathcal{L}(C^\infty(\mathbb T^n))$ there exists $k_A\in C^\infty(\mathbb T^n\times \mathbb T^n)$ such that, $$ Af(x)=\int_{\mathbb T^n} k_A(x, y)f(y)\ dy,$$ for all $f\in C^\infty(\mathbb T^n$)? Thanks

Answer 1

As Dirk hinted, there is no such kernel for the identity operator $Af=f$. Indeed, take $n=1$ for simplicity and consider the sequence $f_m(t)=e^{2\pi mit}$. Integration by parts, $$\int k_A(x,y)e^{2\pi miy}\,dy = -\frac{1}{2\pi mi}\int \frac{\partial k_A(x,y)}{\partial y}e^{2\pi miy}\,dy $$ shows that $\int k_A(x,y)e^{2\pi miy}\,dy\to 0$ pointwise as $m\to\infty$. But this is not the case for $f_m$ itself.

Kernel operators are special, and usually compact (unless the kernel is singular).