# problem 2-23 calculus on manifolds. Help understanding solution

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 $$C_1 = C_2$$

## 1 Answer

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.