# Tensor products of functions generate dense subspace?

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

and

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.

• 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 $Y$. – 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 $$\sup_{(x,y)\in\mathbb R^2}|x^{\alpha_1}y^{\alpha_1}\partial^{\beta}(\varphi(x,y)-\varphi_{n,k}(x,y))|\leq\varepsilon.$$
• 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. – Syang Chen Sep 11 '11 at 0:44
• @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 $C^n$. – Nate Eldredge Sep 13 '11 at 1:15
• @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). – Syang Chen Sep 14 '11 at 2:05
• @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 $P_n(x,y):=Q_n(x,y)+B_n(y)+A-n(x)-\varphi(a_1,a_2)$. – Davide Giraudo Sep 14 '11 at 7:31