Approximation of a two-variable function by tensor products

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$begingroup$
Let $X$ and $Y$ be compact metric spaces and $f: X times Y to mathbb{R}$ be a continuous function.
We know that, for every $n in mathbb{N}$, by the Stone-Weierstrass theorem, there exist $k_n in mathbb{N}$ and continuous functions $f^{(1,n)}_i$, $i in {1, ldots, k_n}$, on $X$ and continuous functions $f^{(2,n)}_i$, $i in {1, ldots, k_n}$, on $Y$, such that
$$ sup_{x in X, y in Y } bigg| f(x,y) - sum_{i=1}^{k_n} f^{(1,n)}_i(x) f^{(2,n)}_i(y) bigg| < frac{1}{n} .$$
Is it possible to choose these functions such that
$$ sup_{n in mathbb{N}} k_n < + infty, $$
$$ sup_{n in mathbb{N}} sup_{i in {1, ldots, k_n}}sup_{x in X} Big| f^{(1,n)}_i(x) Big| < +infty, quad quad sup_{n in mathbb{N}}
sup_{i in {1, ldots, k_n}}sup_{y in Y} Big| f^{(2,n)}_i(y) Big| < +infty quad ? $$
real-analysis functional-analysis approximation-theory
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be compact metric spaces and $f: X times Y to mathbb{R}$ be a continuous function.
We know that, for every $n in mathbb{N}$, by the Stone-Weierstrass theorem, there exist $k_n in mathbb{N}$ and continuous functions $f^{(1,n)}_i$, $i in {1, ldots, k_n}$, on $X$ and continuous functions $f^{(2,n)}_i$, $i in {1, ldots, k_n}$, on $Y$, such that
$$ sup_{x in X, y in Y } bigg| f(x,y) - sum_{i=1}^{k_n} f^{(1,n)}_i(x) f^{(2,n)}_i(y) bigg| < frac{1}{n} .$$
Is it possible to choose these functions such that
$$ sup_{n in mathbb{N}} k_n < + infty, $$
$$ sup_{n in mathbb{N}} sup_{i in {1, ldots, k_n}}sup_{x in X} Big| f^{(1,n)}_i(x) Big| < +infty, quad quad sup_{n in mathbb{N}}
sup_{i in {1, ldots, k_n}}sup_{y in Y} Big| f^{(2,n)}_i(y) Big| < +infty quad ? $$
real-analysis functional-analysis approximation-theory
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be compact metric spaces and $f: X times Y to mathbb{R}$ be a continuous function.
We know that, for every $n in mathbb{N}$, by the Stone-Weierstrass theorem, there exist $k_n in mathbb{N}$ and continuous functions $f^{(1,n)}_i$, $i in {1, ldots, k_n}$, on $X$ and continuous functions $f^{(2,n)}_i$, $i in {1, ldots, k_n}$, on $Y$, such that
$$ sup_{x in X, y in Y } bigg| f(x,y) - sum_{i=1}^{k_n} f^{(1,n)}_i(x) f^{(2,n)}_i(y) bigg| < frac{1}{n} .$$
Is it possible to choose these functions such that
$$ sup_{n in mathbb{N}} k_n < + infty, $$
$$ sup_{n in mathbb{N}} sup_{i in {1, ldots, k_n}}sup_{x in X} Big| f^{(1,n)}_i(x) Big| < +infty, quad quad sup_{n in mathbb{N}}
sup_{i in {1, ldots, k_n}}sup_{y in Y} Big| f^{(2,n)}_i(y) Big| < +infty quad ? $$
real-analysis functional-analysis approximation-theory
$endgroup$
Let $X$ and $Y$ be compact metric spaces and $f: X times Y to mathbb{R}$ be a continuous function.
We know that, for every $n in mathbb{N}$, by the Stone-Weierstrass theorem, there exist $k_n in mathbb{N}$ and continuous functions $f^{(1,n)}_i$, $i in {1, ldots, k_n}$, on $X$ and continuous functions $f^{(2,n)}_i$, $i in {1, ldots, k_n}$, on $Y$, such that
$$ sup_{x in X, y in Y } bigg| f(x,y) - sum_{i=1}^{k_n} f^{(1,n)}_i(x) f^{(2,n)}_i(y) bigg| < frac{1}{n} .$$
Is it possible to choose these functions such that
$$ sup_{n in mathbb{N}} k_n < + infty, $$
$$ sup_{n in mathbb{N}} sup_{i in {1, ldots, k_n}}sup_{x in X} Big| f^{(1,n)}_i(x) Big| < +infty, quad quad sup_{n in mathbb{N}}
sup_{i in {1, ldots, k_n}}sup_{y in Y} Big| f^{(2,n)}_i(y) Big| < +infty quad ? $$
real-analysis functional-analysis approximation-theory
real-analysis functional-analysis approximation-theory
edited Jan 11 at 7:41
Richard
asked Jan 10 at 17:55
RichardRichard
1,2081724
1,2081724
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