When looking at the tensor product of the ring of smooth functions on $\mathbb{R}^n$, there is only an injection $$ C^\infty(\mathbb{R}^n)\otimes_\mathbb{R}C^\infty(\mathbb{R}^m) \to C^\infty(\mathbb{R}^{n+m}) $$ This motivates the construction of the completed tensor product which gives an isomorphism. What is an example of a smooth function which in $$ C^\infty(\mathbb{R}^2) $$ which does not lie in the standard tensor product?

  • $\begingroup$ How about $f(x,y) = x + y$? I think any function which is not a product is an example. $\endgroup$ – Hwang Apr 21 '17 at 2:01
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    $\begingroup$ Why isn't that $x\otimes 1 + 1 \otimes y$? $\endgroup$ – 54321user Apr 21 '17 at 2:05
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    $\begingroup$ I am sorry. I was careless. $\endgroup$ – Hwang Apr 21 '17 at 2:07
  • $\begingroup$ So how about $f(x,y)=\sin(x+y)$? $\endgroup$ – Amitai Yuval Apr 21 '17 at 3:38
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    $\begingroup$ $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y) = \sin(x)\otimes\cos(y) + \cos(x)\otimes\sin(y)$ $\endgroup$ – 54321user Apr 21 '17 at 3:47

Assume that $H \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ is a function that has the form $H(x,y) = \sum_{i=1}^k f_i(x)g_i(y)$ for some $k \in \mathbb{N}$ and some functions $f_i,g_i \colon \mathbb{R} \rightarrow \mathbb{R}$. Then for each fixed $x_0 \in \mathbb{R}$ the function $y \mapsto H(x_0,y)$ is a linear combination of the functions $g_1, \dots, g_k$ (with coefficients in $\mathbb{R}$). In particular, for any $n > k$ the functions

$$ y \mapsto H(1,y), y \mapsto H(2,y), \dots, y \mapsto H(n,y) $$

must be linearly dependent (because they all belong to $\operatorname{span} \{ g_1, \dots, g_k \}$).

So consider for example $H(x,y) = e^{xy}$ and assume that $H = \sum_{i=1}^k f_i(x)g_i(y)$ for some $k$ and $f_i,g_i$. It is readily seen that the functions

$$ e^{y}, e^{2y}, \dots, e^{ky}, e^{(k+1)y} $$

are linearly independent over $\mathbb{R}$ and we arrived a contradiction.


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