Separation of variables If I have a function $f(x,y)$, is it always true that I can write it as a product of single-variable functions, $u(x)v(y)$?
Thanks.
 A: No, it is not.  It is unusual that you can do so.  For example, $f(x,y)=x+y$ cannot be.
A: Unfortunately, the answer is no.  For example, let $f(x,y)=x-y$.
Note that $f(1,1)=0$.  If $f(x,y)=u(x)v(y)$ for all $x$, $y$, then $0=f(1,1)= u(1)v(1)$. But if $u(1)v(1)=0$, then $u(1)=0$ or $v(1)=0$. 
Suppose for example that $u(1)=0$.  Then $u(1)v(y)=0$ for all $y$. If $f(x,y)=u(x)v(y)$ for all $x$, $y$, then $1-y=0$ for all $y$. This is clearly not true.  
The example $f(x,y)=x-y$ is not really special.  "Most" functions $f(x,y)$ are not expressible as $u(x)v(y)$.
A: It is true that linear combinations of such functions are dense. Depending on exactly what space of functions you're working with, this should follow from the locally compact form of the Stone-Weierstrass theorem. Thus it is not unreasonable to expect that we can find solutions to a linear PDE by taking the closure of the space of linear combinations of separable solutions. 
A: If $f(x,y)=x^2+y^2$ could be written as $u(x)v(y)$, then since $f(0,0)=0$, this means either $u(0)=0$ or $v(0)=0$.  If this were the case, $f(x,y)$ would be equal to zero on either the entire $y$-axis or the entire $x$-axis.  This clearly is not the case, so $f(x,y)$ can not be written as $u(x)v(y)$.
You might try graphing a few functions that are of the form $u(x)v(y)$ and see what the graph looks like.  This might give some insight to why the statement is not true.
A: If a multivariable function $f(x_1,\dots,x_n)$ is separable then the Hessian $H\ln f$ is diagonal, which isn't usually the case. Moreover, if $f$ decomposes into $g_1(x_1)\cdots g_n(x_n)$, then
$$\int \frac{\partial\ln f}{\partial x_i} dx_i=g_i(x_i)+C$$
where the integration is done in a single-variable sense.

Can someone prove a $C^2$ function $f$ is always separable when $H\ln f$ is diagonal?
