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This question arises in Zorich, Analysis 1.

If we consider some function $\;$ $f : G \subset \mathbb{R}^2 \rightarrow \mathbb{R}$ $\in$ $C^{(1)}(G;\mathbb{R})$ where the partial derivative is identically zero $\forall (x,y) \in$ $G$, is it possible to assert that the function is independent of $y$?

This seems like a relatively straight forward answer, but inherently produces some subtlety which I shall highlight. That is, we see that in the the direction of the second variable the partial derivative and thus rate of change of the function is zero. Consequently, this must mean that the function values do not vary according to the input of said variable. Now this may indicate some independence of the function in this domain, however I have a doubt that this is true based on the produced counter-example.

Take $\; f: G \subset \mathbb{R}^2 \rightarrow \mathbb{R}$ where $G = \{(x,0) \;| \;x \in \mathbb{R} \}$ and $\; f(x,y) = x^2 + y^2$, then $\partial{f}/\partial{y}|_{G} \equiv 0$. This function is not independent of $y$, yet satisfies the vanishing condition of its partial derivative throughout the domain.

Does this suffice to show that just because the partial derivative of some variable is zero everywhere, the function need not be independent of the variable? Or, have I produced a specific condition on the domain (namely, having the domain be $\mathbb{R} \times \{t\} \subset \mathbb{R}^2$, for some fixed $t \in \mathbb{R}$) which allows for negation of the assertion?

Finally, can we produce a condition on $G$ in which we can definitively state that there is an independence?

Thank you in advance for your responses!

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  • $\begingroup$ I can not understand your example. When you restrict the function $f$ to the domain $G$, $f(x,y)=x^2$ and so the function is independent of the variable $y$ in this domain! $\endgroup$ – Hesam Nov 19 '16 at 7:03
  • $\begingroup$ Thanks for your comment. This is precisely why I have chosen the domain to be this way. After evaluation, the function becomes $f(x,0) = x^2$, yet prior to evaluation the function is explicitly defined as $f(x,y) = x^2 + y^2$, whose partial derivative is $\partial{f} / \partial{y} = 2y$ and thus throughout G vanishes. This may seem pedantic, but I was trying to extract some contradiction of independency using only manipulation of the domain of the function. Can you think of a better counter-example? $\endgroup$ – D.S. Nov 19 '16 at 7:15
  • $\begingroup$ I'd say that in your example the partial derivative with respect to $x$ isn't even defined because we can't take a limit that defines it (which would be a limit as $y\to0$, but the function isn't defined for any $y\neq0$). Maybe there was some kind of openness requirement on the domain? $\endgroup$ – zipirovich Nov 23 '16 at 0:43

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