5
$\begingroup$

how can I show that if $A:C^\infty(\mathbb T^n)\rightarrow C^\infty(\mathbb T^n)$ is a continuous linear operators then there is a unique linear and continuous operator $K_A: C^\infty(\mathbb T^n\times \mathbb T^n)\rightarrow \mathbb C$ such that, $$\langle A\varphi, \psi\rangle=\langle K_A, \psi\otimes \varphi\rangle.$$ Remember that if $u\in C^\infty(\mathbb T^n)$ then $u$ induces a distribution by the pairing: $$\langle u, \phi\rangle=\int_{\mathbb T^n} u(x)\phi(x)\ dx.$$ Indeed, I don't know if this result holds, that would be a version of the Schwartz kernel theorem in the case the distributions are induced by smooth functions... Can anyone help me...

$\endgroup$
4
$\begingroup$

It helps to know that $C^\infty(\mathbf{T}^n)$ can be equivalently topologized by the Sobolev norms $\| \cdot \|_{H^m}$ where $$\|f\|_{H^m}^2 := \sum_{\xi \in \mathbf{Z}^n} (1 + |\xi|)^{2m} |\hat{f}(\xi)|^2,$$ and $\hat{f}(\xi) = \int_{\mathbf{T}^n} e^{-i\xi x} f(x) \, dx$ are the Fourier coefficients of $f$. Indeed, for any $k \ge 0$ and for any integer $\ell$ strictly greater than $n/2$, one has $$ \|f\|_{C^k} \lesssim_k \|f\|_{H^{k + \ell}} $$ and $$ \|f\|_{H^k} \lesssim_k \|f\|_{C^{k + \ell} }. $$ In particular, $|\langle Af, g \rangle|\lesssim \|f\|_{H^n} \|g\|_{H^n}$ (say) for all $f, g \in C^\infty$.

Write $\mathbf{e}_\xi (x) = e^{i \xi x}$. Any $h \in C^\infty(\mathbf{T}^n \times \mathbf{T}^n)$ has a unique Fourier series representation $$h = \sum_{j, k} \langle h, \mathbf{e}_j \otimes \mathbf{e}_k \rangle \mathbf{e}_j \otimes \mathbf{e}_k;$$ by the equivalence of norms, this series converges in $C^\infty$. Thus $K_A$ must have the form $$ K_A(h) = \sum_{j, k} \langle h, \mathbf{e}_j \otimes \mathbf{e}_k \rangle \langle A \mathbf{e}_j, \mathbf{e}_k \rangle,$$ and this is in fact the desired $K_A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.