I am having some trouble understanding the process for proving the following statement:
If $f'(x,v) = 0$ is a closed ball $B(x_0,r)$ for all $x$, show that $f$ is constant.
In the proof, we take two elements $x_1, x_2$ in $B(x_0,r)$, then we define the vector $v$ between $x_1$ and $x_2$ as $v = x_2 – x_1$, so $x_2 = v + x_1$. Finally, we apply the mean value theorem so that:
$f(x_1 +v) -f(x_1) = f'(x_1 + \theta v, v) = 0$
I can not see how the above relates to the mean value theorem. Don't we need to divide by the distance between $x_2$ and $x_1$ on the left hand side? Could someone help me to fill in the missing steps?
Thanks for your help.
Best Answer
$\newcommand{\bm}[1]{\boldsymbol {#1}}$ Let $F(t) = f(\bm x_1 + t \bm v)$ where $t\in [0,1]$. Then $F(0)=f(\bm x_1), F(1)= f(\bm x_2)$ and $F$ is continuous on $[0, 1]$, since each $f'(\bm y, \bm v)$ exists and equals $0$. Now check the differentiability of $F$: for each $k\in (0,1)$, $$ \lim_{h\to 0} \frac {F(k+h) - F(k)}h = \lim_{h\to 0} \frac {f((\bm x_1 +k\bm v) + h \bm v)-f(\bm x_1 + k\bm v)}h, $$ and note that $$ f'(\bm y, \bm v) := \lim_{t\to 0}\frac {f(\bm y + t\bm v) - f(\bm y)}t $$ provided that the limit exists, we have $$ \lim_{h \to 0}\frac {F(k+h) - F(k)}h = f'(\bm x + k\bm v, \bm v), $$ i.e. $$ F'(k) = f'(\bm x_1 +k\bm v , \bm v). $$ So by Lagrange's MVT, $$ F(1) - F(0) = 1 \cdot F'(k), $$ which is just $$ f(\bm x_2) - f(\bm x_1) = f'(\bm x_1 + k\bm v, \bm v), $$ as shown in the OP.
To conclude, that "division by $\bm x_1, \bm x_2$" is included in the expression of the directional derivatives.