I asked the following question a few days ago: Approximation of a $C^{\infty}_c$ function by tensor products However, I then realised that I actually need a stronger result in my proof.

As in the previous question, I suppose that $f \in C^{\infty}_c ( \mathbb{R}^2 )$, i.e. $f$ is a $C^{\infty}$ function with compact support defined on $\mathbb{R}^2$. We know that $f$ can be approximated by a sequence of functions $(f_n)$ in $C^{\infty}_c ( \mathbb{R}) \otimes C^{\infty}_c ( \mathbb{R})$ in the infinity norm, where each $f_n$ is of the form

$$ f_n( x,y) = \sum_{i=1}^{k_n} a^{(n)}_i (x) b^{(n)}_i (y), $$

for some functions $a^{(n)}_i, b^{(n)}_i \in C^{\infty}_c ( \mathbb{R})$, $1 \leq i \leq k_n$.

$\textbf{However, I am interested in fixing the tensor ranks}$ $\{k_n \}_{n \in \mathbb{N}}$ $\textbf{to be a constant}$ $K$, $\textbf{i.e.}$ $k_n = K$, $\textbf{for each}$ $n$.

Therefore, I first wonder whether it is possible to construct such a sequence of functions $(f_n)_{n \in \mathbb{N}}$ which approximates $f$ in the infinity norm AND satisfies this property of constant tensor rank $K$.

If yes, can we further require $(f_n)$ to satisfy the requirement of uniform bound on the $C^1$ norms of the component functions of $x$ and $y$, as in the previous question, i.e. $$ \sup_{n \in \mathbb{N}, 1 \leq i \leq K} \big\| a^{(n)}_i \big\|_{\infty} < + \infty, \quad \quad \sup_{n \in \mathbb{N}, 1 \leq i \leq K} \big\| b^{(n)}_i \big\|_{\infty} < + \infty, $$ and $$ \sup_{n \in \mathbb{N}, 1 \leq i \leq K} \big\| \big( a^{(n)}_i \big)' \big\|_{\infty} < + \infty, \quad \quad \sup_{n \in \mathbb{N}, 1 \leq i \leq K} \big\| \big( b^{(n)}_i \big)' \big\|_{\infty} < + \infty?$$

I actually found out that there is a wide range of literature on the topic of approximation by tensor products with constant tensor ranks, e.g.




In particular, the second link (the paper by Temlyakov) gives $L^p$ estimates of the error incurred by approximation with tensor products with constant tensor rank. However, I can't find any paper that addresses the aforementioned issue of a uniform bound on the $C^1$ norms of the component functions. Since I am not in the area of approximation theory, any help/references would be greatly appreciated!

  • 1
    $\begingroup$ Not sure if I am mistaken, but if you evaluate $ f$ on a matrix of $K+1$ by $K+1$ points and get nonzero determinant, then approximation by rank $K$ tensor is impossible, even in the sup norm, as all rank $K$ tensors evaluated on such a matrix of points will give a matrix of rank(!) at most K. $\endgroup$
    – Fan Zheng
    Nov 29, 2016 at 2:52

1 Answer 1


Fixing the rank of the approximating tensors yields closed subsets: More precisely: Given two Banach spaces $X$ and $Y$ and $n\in\mathbb N$ the set $\otimes_n(X,Y)=\lbrace \sum_{i=1}^n x_i\otimes y_i: x_i\in X, y_i\in Y\rbrace$ is closed in $X\tilde{\otimes}_\varepsilon Y$. (This is exercise 12.7 in the book Tensor Products and Operator Ideals of Defant and Floret, they refer to an article of Valdivia A class of locally convex spaces without C-webs, Ann. Inst. Fourier 32 (1982), 261-269). Since the $\varepsilon$-tensor product for spaces of continuous functions corresponds to uniform approximation this should give a negative answer to your question.


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