Continue with my previous question “Regarding Kolmogorov's Superposition Theorem”, here are some further questions:

Question-1 Is it true: for any given $C^1$ continuous real function $f(x, y): \Re^2 \to \Re^1$, there exist $C^1$ continuous real functions $\Psi(x), w(x,y), u(x)$ such that: $u(f(x,y))=u(x)\Psi(\frac{w(x, y)}{u(x)})$ where $\frac{du(x)}{dx}\ge 0$; $\frac{\partial w(x, y)}{\partial y}\ge 0$

Question-2 Can these functions $\Psi(x), w(x,y), u(x)$ be solved, analytically or approximately?

It is found that constraint $\frac{\partial w(x,y)}{\partial y}\ge 0$ is hard to deal with. Without this condition, there is one obvious solution $u(x)=x, w(x, y)=f(x,y), \Psi(x)=x$, where $\frac{du(x)}{dx}=1>0$, but $\frac{\partial w(x, y)}{\partial y}\ge 0$ may not satisfy.

One may find by letting $w(x,y)=w(y)$, this becomes the question we asked in "Regarding Kolmogorov's Superposition Theorem" but since the answer is NO found by AgCl, we want to remove the constraint $w(x,y)=w(y)$ hoping solution exist...

Continue with my previous question “Regarding Kolmogorov's Superposition Theorem”, here are some further questions:

Question-1 Is it true: for any given $C^1$ continuous real function $f(x, y): \Re^2 \to \Re^1$, there exist $C^1$ continuous real functions $\Psi(x), w(x,y), u(x)$ such that: $u(f(x,y))=u(x)\Psi(\frac{w(x, y)}{u(x)})$ where $\frac{du(x)}{dx}\ge 0$; $\frac{\partial w(x, y)}{\partial y}\ge 0$

Question-2 Can these functions $\Psi(x), w(x,y), u(x)$ be solved, analytically or approximately?

It is found that constraint $\frac{\partial w(x,y)}{\partial y}\ge 0$ is hard to deal with. Without this condition, there is one obvious solution $u(x)=x, w(x, y)=f(x,y), \Psi(x)=x$, where $\frac{du(x)}{dx}=1>0$, but $\frac{\partial w(x, y)}{\partial y}\ge 0$ may not satisfy.

One may find by letting $w(x,y)=w(y)$, this becomes the question we asked in "Regarding Kolmogorov's Superposition Theorem" but since the answer is NO found by AgCl, we want to remove the constraint $w(x,y)=w(y)$ hoping solution exist...

Hi AgCl and Experts,

Continue with my previous question “Regarding Kolmogorov's Superposition Theorem”, here are some further questions:

Question-1 Is it true: for any given $C^1$ continuous real function $f(x, y): \Re^2 \to \Re^1$, there exist $C^1$ continuous real functions $\Psi(x), w(x,y), u(x)$ such that: $u(f(x,y))=u(x)\Psi(\frac{w(x, y)}{u(x)})$ where $\frac{du(x)}{dx}\Ge 0$; $\frac{\partial w(x, y)}{\partial y}\Ge 0$

Question-2 Can these functions $\Psi(x), w(x,y), u(x)$ be solved, analytically or approximately?

It is found that constraint $\frac{\partial w(x,y)}{\partial y}\Ge 0$ is hard to deal with. Without this condition, there is one obvious solution $u(x)=x, w(x, y)=f(x,y), \Psi(x)=x$, where $\frac{du(x)}{dx}=1>0$, but $\frac{\partial w(x, y)}{\partial y}\Ge0$ may not satisfy.

One may find by letting $w(x,y)=w(y)$, this becomes the question we asked in "Regarding Kolmogorov's Superposition Theorem" but since the answer is NO found by AgCl, we want to remove the constraint $w(x,y)=w(y)$ hoping solution exist...

Thank you!! Wang Tao

Continue with my previous question “Regarding Kolmogorov's Superposition Theorem”, here are some further questions:

Question-1 Is it true: for any given $C^1$ continuous real function $f(x, y): \Re^2 \to \Re^1$, there exist $C^1$ continuous real functions $\Psi(x), w(x,y), u(x)$ such that: $u(f(x,y))=u(x)\Psi(\frac{w(x, y)}{u(x)})$ where $\frac{du(x)}{dx}\ge 0$; $\frac{\partial w(x, y)}{\partial y}\ge 0$

Question-2 Can these functions $\Psi(x), w(x,y), u(x)$ be solved, analytically or approximately?

It is found that constraint $\frac{\partial w(x,y)}{\partial y}\ge 0$ is hard to deal with. Without this condition, there is one obvious solution $u(x)=x, w(x, y)=f(x,y), \Psi(x)=x$, where $\frac{du(x)}{dx}=1>0$, but $\frac{\partial w(x, y)}{\partial y}\ge 0$ may not satisfy.

One may find by letting $w(x,y)=w(y)$, this becomes the question we asked in "Regarding Kolmogorov's Superposition Theorem" but since the answer is NO found by AgCl, we want to remove the constraint $w(x,y)=w(y)$ hoping solution exist...