Let $f$ be a scalar field, $f:\mathbb R^n\to\mathbb R$. Suppose there is an $n$-ball $B(a;r)$ centered at $a$ with radius $r$ and a fixed vector $y\in\mathbb R^n$ such that $f'(x;y)=0$ for every $x\in B(a;r)$. What can you conclude about $f$?
This problem is from Apostol Calculus Volume 2. I have just begun the subject. However, I did try to work out as follows:
It is evident, from Mean Value Theorem, that for any $x\in B(a;r)$, we have $f(x+y)-f(x)=f'(x+\theta y;y)=0$, supposing that $x+y\in B(a;r)$ and hence $x+\theta y\in B(a;r)$ if $0<\theta<1$. (Here's my first question: can we assume that just because $x\in B(a;r)$ that $x+y\in B(a;r)$ also?)
Anyway this gives that $f(x+y)=f(x)$ for each $x\in B(a;r)$.
I am not sure what I am supposed to conclude from here. I tried to interpret it by drawing an open ball around $a$ and taking a few points $x_1,x_2,x_3$ inside the ball, I tried to plot $x_i+y$ (in parallel straight lines), $i=1,2,3$ but I couldn't understand if this is anything significant.
Any help will be appreciated.
The symmetry $f(x+ty)=f(x)$ (whenever $x,x+ty\in B(a;r)$) is in fact equivalent with $f'(x;y)=0$ in $B(a;r)$. You can observe this by differentiating in the direction of $y$; you need not consider any other derivatives.
If you write $\mathbb R^n=\mathbb R\oplus y^\perp$ (directions parallel and orthogonal to $y$), the function $f$ only depends on the variable in $y^\perp$ (which is $n-1$ dimensional). If $y=(1,0,\dots,0)$, the conclusion is clearer. Then $f$ only depends on $x_2,\dots,x_n$, not on $x_1$.
It does not follow from $x\in B(a;r)$ that also $x+y\in B(a;r)$. But since the ball is open, it does follow that for real numbers $t$ close enough to zero, $x+ty\in B(a;r)$. You can use the mean value theorem (or the fundamental theorem of calculus along a line) on shorter line segments just as well.