Assume that $f′(x;y)=0$ for every $x$ in some $n$-ball $B(a)$ and for every vector $y$.

Use the mean value theorem to prove that $f$ is constant on $B(a)$.

And if $f′(x;y)=0$ for a fixed vector $y$ and for every $x$ in $B(a)$, what can you conclude about $f$ in this case?


2 Answers 2


Let $\bar{x} \in B(\bar{a})$ and $ \bar{y} \in \mathbb{R}^n $.

Note that $B(\bar{a})$ is an open set. So we can find an $r>0$ such that $B(\bar{x},r) \subseteq B(\bar{a})$. Hence, we can find a $\theta >0$ such that $\bar{x}+h\bar{y} \in B(\bar{x},r),$ for all $ h \in [-\theta,\theta]$. (I request you to make a diagram to convince yourself about this argument.)

Now define the function $g(t) = f(\bar{x}+t\bar{y})$. Now apply one-dimensional mean-value theorem to $g$ on the interval $[-\theta,\theta]$ :

For any given $h \in [-\theta,\theta],$ there exists a $\xi \in (-\theta,\theta)$ such that $\frac{g(h)-g(0)}{h} = g'(\xi)$.

As $g'(\xi) = f'(\bar{x}+\xi\bar{y};\bar{y}) = 0$, we obtain $g(h) = g(0) \Rightarrow f(\bar{x}+h\bar{y}) = f(\bar{x})$, for all $h \in [-\theta,\theta]$.

Note that $g$ and $h$ depend on our choice of $\bar{x}$ and $\bar{y}$. But these were arbitrary and hence we are done.

For the second part, $f$ is constant as we move in the direction of $\bar{y}$ across the ball $B(\bar{a})$.


We know $f'(x;y)=o$ for every x in n-ball B(a) i.e x such that $\Vert x-a \Vert \lt r$

For $x \in B(a)$ let $y= x-a$ and $z(t)= tx + (1-t)a, 0 \le t \le 1$

  1. $z(t) \in B(a) $ since n-ball is convex

    therefore $f'\left( z(t);y\right) = 0$

  2. By mean-value theorem $$f(a+y) - f(a) = f'\left( z(t'\right);y) =0 $$ for some $t'$ s.t $ 0\le t' \le1 $ $$f(a+y) = f(x)=f(a) $$ and since $x \in B(a)$ was arbitarly taken, therefore $f(x)=f(a)$ for every $ x \in B(a)$ \line Hence f is constant


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