Assume $f$ is differentiable at each point of an n-ball $B(a)$. Prove that if $f(x) \leq f(a)$ for all $x$ in $B(a)$, then $\nabla {f(a)} = 0.$

I had my proof, but I'm not sure it is correct.

Proof: Since f is differentiable at each point of the n-ball B(a), meaning

$$\lim_{h \to 0} \frac{f(a+hy)-f(a)}{h} = \nabla f(a) \cdot y$$ , where y is an arbitrary unit vector.

From the mean value theorem, we know that

$$\lim_{h \to 0} \frac{f(a+hy)-f(a-hy)}{h} = \nabla f(c) \cdot y$$ for some c where $||c|| < r$.


$$\lim_{h \to 0} \frac{f(a+hy)-f(a)}{h} = \nabla f(c) \cdot y = - \lim_{h \to 0} \frac{f(a)-f(a-hy)}{h}$$

Since the RHS of the the first equation is 0, we have $\nabla f(a) = 0$.

So, is there any mistake of any suggestion about the point that I can improve mathematically or about the way that I wrote ?

  • $\begingroup$ Rather use the symbol \nabla. $\endgroup$ – Did Nov 5 '16 at 6:30
  • $\begingroup$ @Did thanks for pointing out $\endgroup$ – onurcanbektas Nov 5 '16 at 7:27

You never used the fact that $f(x)\leq f(a)$. I may be missing something, but it seems that your proof would imply that all differentiable functions have this property.

My proof is the following: Compute

$$\lim_{h\to 0 } \frac{f(a+hy)-f(a)}{h} = \nabla f(a) \cdot y,$$

and note that $f(a+hy)\leq f(a)$ for all $h$ sufficiently small, so $\nabla f(a)\cdot y\leq 0$. However, taking $y\mapsto -y=:\tilde{y}$ we again have

$$\lim_{h\to 0} \frac{f(a+h\tilde{y})-f(a)}{h} = \nabla f(a)\cdot \tilde{y},$$

implying $\nabla f(a) \cdot \tilde{y} \leq 0$. But, since $\tilde{y}=-y$ we have $\nabla f(a)\cdot y \geq 0$. Combining this with the first inequality we derive $\nabla f(a) \cdot y =0$. Since $y$ was arbitrary it immediately follows that $\nabla f(a)=0.$

EDIT: To make the last argument explicit: Our above work implies that $\nabla f(a) \cdot y =0 $ for any $y$. In particular, take $y=\nabla f(a)$ so that

$$0 = \nabla f(a) \cdot \nabla f(a) = |\nabla f(a)|^2.$$

Then, $|\nabla f(a)|^2 = 0$ only if $\nabla f(a) =0.$

  • $\begingroup$ I couldn't see the combination in the last sentence, could you express it explicitly ? $\endgroup$ – onurcanbektas Nov 5 '16 at 7:26
  • $\begingroup$ @user251257 it didn't help at all :) $\endgroup$ – onurcanbektas Nov 5 '16 at 11:53
  • $\begingroup$ @Leth I have made an edit to the proof so that the conclusion should be clear. Please let me know if there is any confusion. $\endgroup$ – Matt Nov 5 '16 at 14:28

Let $y \in R^n$ an arbitrary vector and let be $\alpha: (-\epsilon,\epsilon) \to B(a)$ given $\alpha(h) = a + hy$. Then the differentiable function $g: (-\epsilon,\epsilon) \to R$ where $g = f \circ \alpha$ has critical value in $h=0$ and therefore $$ 0 = g´(0) = \nabla f . y$$, how $y$ is arbitrary, this implies that $\nabla f = 0$

  • $\begingroup$ how do we know that g has a critical value at h=0 ? $\endgroup$ – onurcanbektas Nov 5 '16 at 8:43
  • $\begingroup$ @Leth It has a local minimum and is differentiable at $0$. $\endgroup$ – egreg Nov 5 '16 at 9:46
  • $\begingroup$ Can you prove it ? Without proving saying that it has some property is meaningless unless it is obvious. $\endgroup$ – onurcanbektas Nov 5 '16 at 9:54
  • $\begingroup$ Your Hipotesis that $f(x) \leq f(a)$ more the fact that I'm do choosing $\epsilon > 0$ such that $\alpha(h) \in B(a)$ for for all $|h| < epsilon$ implies that $g$ has maximal value in $h = 0$ and therefore $h =0$ is critical point. $\endgroup$ – A.D. Nov 5 '16 at 10:49
  • $\begingroup$ I'm sorry I can't see the obvious implication, but I'm lost with the phrase "implies".I guess I don't remember the theorem that allows you to say "implies" $\endgroup$ – onurcanbektas Nov 5 '16 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.