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?

  • How about $f(x,y) = x + y$? I think any function which is not a product is an example. – Hwang Apr 21 '17 at 2:01
  • 1
    Why isn't that $x\otimes 1 + 1 \otimes y$? – 54321user Apr 21 '17 at 2:05
  • 1
    I am sorry. I was careless. – Hwang Apr 21 '17 at 2:07
  • So how about $f(x,y)=\sin(x+y)$? – Amitai Yuval Apr 21 '17 at 3:38
  • 2
    $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y) = \sin(x)\otimes\cos(y) + \cos(x)\otimes\sin(y)$ – 54321user Apr 21 '17 at 3:47
up vote 7 down vote accepted

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.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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