Let $A = \{(x,y): x<0, \text{ or } x \geq 0 \text{ and } y\neq 0 \}$

If $f:A \to R$ and $D_1 f = D_2 f = 0$ show that $f$ is constant

I found the following solution here http://www.vision.caltech.edu/~kchalupk/spivak.html but I don't understand it.

The solution reads

Let $(x_1,y_1)$ and $(x_2,y_2)$ be two points in $A$. Because $D_1f=0$, the value of $f$ is constant along the lines $l_1=\{(x,y_1)|x \in R\}$ and $l_2=\{(x,y_2)|x \in R\}$. In particular, we have $f(−1,y_1) = f(x_1,y_1)$, $f(−1,y_2)=f(x_2,y_2)$. Because $D_2 f = 0$ we must also have $f(−1,y_1)=f(−1,y_2)$, hence $f$ is constant on $A$

What I don't understand is why $f(−1,y_1)=f(−1,y_2)$ follows from $D_2 f = 0 $. If $D_2 f = 0$ Independence on the second variable comes from the mean value theorem.

The mean value theorem requires continuity on $[y_1, y_2]$ but the function is not defined for $y = 0$, so if $y_1 < 0$ and $y_2 > 0$ we can't use the mean value theorem because $f$ is not continuous on $[y_1, y_2]$. if $y_1, y_2 < 0$ then we have $f(x,y_1) = f(x,y_2) = C_1$ and if $y_1,y_2 > 0$ we have $f(x,y_1) = f(x,y_2) = C_2$ but we can't assume that $C_1 = C_2$


I think you're not understanding the definition of the set $A$ correctly. The set is

$$ A = \{(x,y) \in \mathbb R^2: x < 0, \text { or } x \geq 0 \text { and } y \neq 0 \}.$$

You should be reading the conditions defining $A$ as $\big(x< 0 \big) \text { or } \big( x \geq 0 \text { and } y \neq 0 \big).$ So the set $A$ is simply the entire plane $\mathbb R^2$ minus the line $\{(x,0): x \geq 0 \}$.

Therefore the line segment joining $(-1,y_1)$ and $(-1,y_2)$ always lies in $A$ and $f$ is well defined on this line. You can use the mean value theorem to conclude that $f$ is constant on this line segment.


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