If $f$ is a real function defined in a convex open set $E \subset \mathbb{R}^n$ such that the partial derivative of $f$ with respect to, say, $x_1$ is identically $0$, we want to show that $f$ depends on ly on the other variables.

It is easy to do this proof using the convexity hypothesis. We can write down two points $f(x_1', ..., x_n)$ and $f(x_1'', x_2, ... x_n)$, and draw the line between them. This line lies entirely in $E$. The Mean Value Theorem applies on this line, and implies that there is some point in the interior $c$ so that the $x_1$ derivative at $c$ is equal to $$\frac{f(x_1'', x_2, ..., x_n) - f(x_1', x_2, ..., x_n)}{x_1'' - x_1'}$$ But this derivative is $0$, and so it follows $f$ is independent of $x_1$.

Rudin continues, asking the reader to find a weaker condition which suffices, and to find examples showing that some condition is necessary. I think for this last part, we can do something simple enough. We can maybe take a horizontal slice out of the plane, and define a function which is $0$ on the bottom half and left side of the plane, but in the first quadrant varies with $y$. We can probably do this in a way which is continuous and differentiable, I just haven't thought about it quite long enough yet.

Anyway, the real question that has me stuck is what the weaker condition should be. If I had to take a guess, I'd guess star-shapedness, but I seem to be struggling to prove this. Or maybe it's wrong all together.

  • $\begingroup$ I would guess that the sets $E_{ (x_2,...,x_n) } = \{x_1 | (x_1,...,x_n) \in E \}$ should be intervals. $\endgroup$ – copper.hat May 8 '17 at 3:42
  • $\begingroup$ That seems like a good idea. I'll try that. $\endgroup$ – Alfred Yerger May 8 '17 at 3:44

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