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If $G$ is an open connected set in $\mathbb R^2$ and $f$ is continuous function on $G$ such that the partial derivatives vanish identically. Then show that $f$ is constant on $G$ .

How can I use the connectedness of $G$ here to solve this any help ?

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    $\begingroup$ Continuity of the partial derivatives implies total differentiability. $\endgroup$ – Friedrich Philipp Mar 21 '16 at 11:21
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    $\begingroup$ Hint: f(G) is connected too $\endgroup$ – Martín Vacas Vignolo Mar 21 '16 at 11:34
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The partial derivatives are constant, hence differentiable and $f$ is differentiable as well. Since the partial derivatives vanish, $f$ is locally constant! Here you need the connectedness to conclude that $f$ is constant.

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In $\;\Bbb R^n\;$ , open and connected gives us path connected. With this I think we can already begin. The details I'll let you to fill.

Take any point $\;a\in G\;$ , and now take a point $\;u\in A\;$ such that the straight line $\;r(t):=tu+(1-t)a\;,\;\;t\in[0,1]\;$ is wholly contained in $\;G\;$ . This is possible for any $\;u\in G\;$ if $\;G\;$ is convex, for example, and anyway it is always possible by means of a polygonal line which will be almost always differentiable. You will need to complete in this last case.

Let us define now

$$\;g(t):=f(r(t))\implies g'(t)=f'(r(t))\bullet r'(t)$$

But $\;f'(r(t))=\nabla (f(r(t))=\vec0\;$ since all the partial derivatives are zero, so the above dot (inner) product is zero, which means $\;g(t)=0\iff g(t)\;$ is a constant, and then

$$g(1)=f(r(1))=f(u)=f(a)=f(r(0))=g(0)\implies f(u)=f(a)$$

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