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Thats maybe a stupid question, it is really trivial, but I asked me how one could really show this.

If $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is a differentiable function and $\frac{\partial f}{\partial y}=0$ then f does not depend on y.

I mean it is clear, but how can we show this, because when I use contradiction and the definition of the partial derivative I somhow get stuck at $$\frac{\partial f}{\partial y}(x,y)=\lim_{h\rightarrow 0} \frac{f((x,y+h))-f(x,y)}{h}$$

Thank you for your help.

user1294729
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    Try the mean value theorem. Suppose that $f(x_0,y_0)\neq f(x_0, y_1)$ and derive a contradiction. – lulu Dec 16 '21 at 22:36
  • sorry I do not see it yet – user1294729 Dec 16 '21 at 22:41
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    Well, try. That hint is very close to a full solution. – lulu Dec 16 '21 at 22:41
  • so but I need to use my approach from above? – user1294729 Dec 16 '21 at 22:42
  • You didn't indicate any approach. All you did was write down the usual definition of a partial derivative. – lulu Dec 16 '21 at 22:43
  • If $f$ does not depend on $y$, then either $f(x,y) = f(x)$ ($f$ depends only on $x$) or $f$ is constant ($f$ depends on neither variable). You should have no trouble applying the definition to these two cases. – Eric Towers Dec 16 '21 at 22:43
  • @lulu so I think that we have to apply to the definition the mean value theorem, but for this theorem doesn't we need the domain to be compact – user1294729 Dec 16 '21 at 22:46
  • The domain is compact. The relevant domain is just the interval $[y_0, y_1]$. Or, if you prefer, the line segment connecting $(x_0, y_0)$ to $(x_0, y_1)$. – lulu Dec 16 '21 at 22:47
  • but this is not the domain of f – user1294729 Dec 16 '21 at 22:48
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    Please try to work the problem yourself. This argument is exactly the same as the usual proof (via the mean value theorem) that a differentiable function of one real variable with derivative $0$ is a constant. If you are unfamiliar with that proof, I suggest starting there. – lulu Dec 16 '21 at 22:49
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    Maybe you may want to write in precise terms what does it mean for f not to depend on y at all. It will then be clear why this relates to lulu's suggestion. – Gollum Dec 16 '21 at 22:52
  • so I understood it as follows: f does not depend on y means that f has no y in the definition – user1294729 Dec 16 '21 at 22:54
  • The mean value theorem applies to functions of multiple variables, if only one variable changes. $$f(x,y_1,z)-f(x,y_2,z)=(y_1-y_2) f_y(x,y_3,z)$$ for some $y_3$ between $y_1,y_2.$ – Thomas Andrews Dec 16 '21 at 22:56
  • Ah, so I can proof it as follows. Let us assume that $f(x,y_0)\neq f(x,y_1)$ so that it does depend on y, we can look at $f:\mathbb{R}\times [y_0,y_1]\rightarrow \mathbb{R}$ Then since f is differentiable, it is continuous on $[y_0,y_1]$. Therefore we can use the mean value theorem, so there exists $y_3\in (y_0,y_1)$ such that $f_{y_3}(x,y_3)=\frac{f(x,y_1)-f(x,y_0)}{(0,y_1-y_0}$ But then by assumption $0=\frac{f(x,y_1)-f(x,y_0)}{(0,y_1-y_0}\Leftrightarrow f(x,y_1)=f(x,y_0)$. But this is a contradiction to the assumption. – user1294729 Dec 16 '21 at 23:07
  • Does this work? – user1294729 Dec 16 '21 at 23:16

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Let's start with the single variable case and then it'll be clear how it applies to multiple variables. The mean value theorem states that if we have a function $f(x)$ that is continuous on an interval $[a,b]$ and differentiable on $(a,b)$, then there is some $c\in (a,b)$ where $$f'(c) = \frac{f(b)-f(a)}{b-a},$$ which essentially means that the "average slope" of a function over an interval must be equal to the instantaneous slope at one point in that interval.

Let's assume that $f\colon [a,b] \to \mathbb{R}$ is a continuous function on $[a,b]$ and differentiable on $(a,b)$ with $f'(x) = 0$ for all $x\in (a,b)$. Striving for contradiction, assume $f(x_1) \ne f(x_2)$, for some $x_1,x_2 \in [a,b]$, i.e. that $f$ is nonconstant. Indeed, then we have $$\frac{f(x_2)-f(x_1)}{x_2-x_1}\ne 0$$ which means that for some $c \in (x_0, x_1)\ne \varnothing$ we have $f'(c) \ne 0$, which contradicts our assumption that $f'(x)$ was identically zero on $(a,b)$.

Generalizing to the multivariable setting, let $f\colon \mathbb{R}^2 \to \mathbb{R}$ be differentiable (and thus continuous on $\mathbb{R}^2$) such that $\frac{\partial f}{\partial y} = 0$. Then, we'll assume that $f$ depends on $y$, i.e. that $f(x, y_1)\ne f(x,y_2)$ for some $y_1\ne y_2$ and strive for contradiction. If we define $g\colon \mathbb{R}\to \mathbb{R},\ g(t) := f(x,t)$, then $g'(t) = 0$ for all $t$ by our assumption. However, $g(y_1) \ne g(y_2)$, so we can apply the above result to this now single variable function and obtain a contradiction.

Sasha Berger
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