Let $X$ and $Y$ be two spaces in certain category, $F(\cdot)$ a functor associating each space with a function space (with certain topology). Assume that for any $f\in F(X)$ and $g\in F(Y)$, $f\otimes g := f(x)g(y)\in F(X\times Y)$. Denote by $F(X)\otimes F(Y)$ the closed subspace of $F(X\times Y)$ generated by functions of the form $f\otimes g$.

General Question: When do we have $F(X\times Y)=F(X)\otimes F(Y)$?

Applying Stone-Weierstrass theorem, we obtain an example:

Theorem: Let $X$ and $Y$ be compact Hausdorff spaces, $C(X)$ and $C(Y)$ the space of continuous functions on X and Y respectively, then we have $C(X\times Y)=C(X)\otimes C(Y)$.

As one can imagine, this type of results could be very useful in reducing multi-dimensional problems into one-dimensional problems. The following are two concrete questions asking for validity of the identity (and they are the true purpose of this post).

Question I: $C^{\infty}_c(\mathbb{R}\times \mathbb{R})=C^{\infty}_c(\mathbb{R})\otimes C^{\infty}_c(\mathbb{R})$?


Question II: $\mathscr{S}(\mathbb{R}\times \mathbb{R})=\mathscr{S}(\mathbb{R})\otimes \mathscr{S}(\mathbb{R})$?

Here $C^{\infty}_c(\mathbb{R})$ denotes the space of smooth functions with compact support, $\mathscr{S}(\mathbb{R})$ denotes the Schwartz space.

  • $\begingroup$ One has also $C_c^{\infty}(X) \otimes C_c^{\infty}(Y) = C_c^{\infty}(X \times Y)$ for all totally disconnected locally compact Hausdorff spaces $X$ and $Y$. $\endgroup$ – m.s Mar 27 '18 at 10:47

Question I: Let $\varphi\in C^{\infty}_c(\mathbb R\times \mathbb R)$. The support of $\varphi$ is contained in $A\times B$ where $A$ and $B$ are two compact subsets of $\mathbb R$. Let $b_1$ and $b_2$ bump function with $b_1=1$ on $A$ and $b_2=1$ on $B$. Let $K_1$ the support of $b_1$ and $K_2$ the support of $b_2$. Thanks to Stone-Weierstrass theorem, we can find a sequence of polynomials $\{P_n\}$ such that $\partial^{\alpha}P_n$ converges uniformly on $K_1\times K_2$ to $\partial^{\alpha}\varphi$ for each $\alpha\in\mathbb N^2$. Now put $\varphi_n(x,y):=b_1(x)b_2(y)P_n(x,y)$. Then $\varphi_n\in C_c^{\infty}(\mathbb R) \otimes C_c^{\infty}(\mathbb R)$ since we can write $$ P_n(x,y) =\sum_{\alpha_1+ \alpha_2\leq d_n}a_{\alpha_1,\alpha_2}b_1(x)x^{\alpha_1}b_2(y)y^{\alpha_2},$$ where $d_n\in\mathbb N$ and $a_{\alpha_1,\alpha_2}\in\mathbb C$. We can check that $\{\varphi_n\}$ converges to $\varphi$ in $C_c^{\infty}(\mathbb R\times \mathbb R)$: $\operatorname{supp} \varphi_n\subset K_1\times K_2$ and the Leibniz rule shows that for each $\alpha \in\mathbb N^2$, the sequence $\{\partial^{\alpha}\varphi_n\}$ converges uniformly on $K_1\times K_2$ to $\partial^{\alpha}\varphi$.

Question II: We show that $C_c^{\infty}(\mathbb R^2)$ is a dense subspace of $\mathcal S(\mathbb R^2)$. Let $\chi\in C_c^{\infty}(\mathbb R^2)$ such that $\chi(u)=1$ if $\lVert u\rVert \leq 1$, and put $\chi_R(x,y):=\chi\left(\frac xR,\frac yR\right)$. Now let $\varphi\in \mathcal S(\mathbb R^2)$. We put $\varphi_R(x,y)= \chi_R(x,y)\varphi(x,y)$, and we can see thanks to the Leibniz rule that $\sup_{(x,y)\in\mathbb R^2}|x^{\alpha_1}y^{\alpha_2}\partial ^{\beta} (\varphi(x,y)-\varphi_R(x,y))|\leq C_{\alpha,\beta}R^{-1}+\sup_{x^2+y^2\geq R}| x^{\alpha_1}y^{\alpha_2}\partial^{\beta}\varphi(x,y)|$, hence that the sequence $\{\varphi_n\}$ converge to $\varphi$ in $\mathcal S(\mathbb R^2)$.

Now, we can, for each $n$, find a sequence $\{\varphi_{n,k}\}_k\subset C_c^{\infty}(\mathbb R)\otimes C_c^{\infty}(\mathbb R)\subset \mathcal S(\mathbb R)\otimes \mathcal S(\mathbb R)$ which converges to $\varphi_n$ in $C_c^{\infty}(\mathbb R^2)$. We fix $\varphi\in \mathcal S(\mathbb R^2)$, $\varepsilon>0$ and $\alpha,\beta\in\mathbb N^2$. We can find a $n$ such that $\sup_{(x,y)\in\mathbb R^2}|x^{\alpha_1}y^{\alpha_1}\partial^{\beta}(\varphi(x,y)-\varphi_n(x,y))| \leq \frac{\varepsilon}2$, and since $\{\varphi_{n,k}\}$ converges to $\varphi_n$ in $C_c^{\infty}(\mathbb R^2)$, we can find a compact subset $K$ of $\mathbb R^2$ such that the sequence $\{\partial^{\beta}\varphi_{n,k}\}$ converges uniformly on $K$ to $\partial^{\beta}\varphi_n$, and since $K$ is compact $$\sup_{(x,y)\in\mathbb R^2}|x^{\alpha_1}y^{\alpha_1}\partial^{\beta}(\varphi_n(x,y)-\varphi_{n,k}(x,y))| \leq \sup_{(x,y)\in K}\lVert (x,y)\rVert \sup_{(x,y)\in\mathbb R^2}|\partial^{\beta}(\varphi(x,y)-\varphi_n(x,y))|.$$ We pick $k$ such that $$\sup_{(x,y)\in\mathbb R^2}|\partial^{\beta}(\varphi(x,y)-\varphi_n(x,y))|\leq \dfrac{\varepsilon}{2\sup_{(x,y)\in K}\lVert (x,y)\rVert},$$ therefore $$\sup_{(x,y)\in\mathbb R^2}|x^{\alpha_1}y^{\alpha_1}\partial^{\beta}(\varphi(x,y)-\varphi_{n,k}(x,y))|\leq\varepsilon.$$

  • 1
    $\begingroup$ but why can you find polynomials $P_n$ such that $P_n$ converge to $\varphi$ in $C^{\infty}$? As far as I know, this can be done only in $C^{k}$ topology where $k$ is any finite natural number. $\endgroup$ – Syang Chen Sep 11 '11 at 0:44
  • 1
    $\begingroup$ @chenn: Choose $P_n$ such that $||\partial_\alpha (P_n - \varphi)||$ is less than $1/n$ for all $|\alpha| \le n$. That is, $P_n$ is close to $\varphi$ in $C^n$. $\endgroup$ – Nate Eldredge Sep 13 '11 at 1:15
  • $\begingroup$ @Nate: Yes you are right. I made a false claim. Now the problem is how to approximate a smooth function defined on a closed rectangle, in $C^1$ topology for example (this could be done easily on an intervel though). $\endgroup$ – Syang Chen Sep 14 '11 at 2:05
  • $\begingroup$ @chenn: We can find a sequence of polynomials $\{Q_n\}$ which converges uniformly on $\left[a_1,b_1\right]\times\left[a_2,b_2\right]$ to $\partial_{xy}\varphi$. Let $P_n(x,y):=\int_{a_1}^x\int_{a_2}^yQ_n(t_1,t_2)dt_2dt_1$, two sequences $\{A_n(x)\}$ and $\{B_n(y)\}$ of polynomials which converges respectively to $x\mapsto \varphi(x,a_2)$ and $y\mapsto\varphi(a_1,y)$ in $C^1$. Now consider $P_n(x,y):=Q_n(x,y)+B_n(y)+A-n(x)-\varphi(a_1,a_2)$. $\endgroup$ – Davide Giraudo Sep 14 '11 at 7:31

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.