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I'm working on a proof on my own, because I didn't like the proofs I found in the books. Can anyone help me checking whether it is correct or not?

Let $f: U\rightarrow \mathbb R$ be a $C^2$ function defined on the open set $U\subset \mathbb R^n$. Suppose $p\in U$ is such that $\nabla f(p)=0$ and

$$\langle Hf(p)\cdot v, v\rangle<0$$ for every $v\in \mathbb R^n\setminus\{0\}$, where

$$Hf(p)=\left(\frac{\partial^2 f}{\partial x_i \partial x_j}(p)\right)$$

is the Hessian matrix of $f$ at $p$. I'd like to show there is $\delta>0$ such that

$$f(x)<f(p)$$ for every $p\in B(x, \delta)\subset U$, that is, $p$ is a local maximum.

Well, since $U$ is open we can find $\delta_1>0$ such that $B(p, \delta_1)\subset U$. But then $f|_{B(p, \delta_1)}: B(p, \delta_1)\rightarrow \mathbb R$ is a $C^2$ function defined on the convex set $B(p, \delta_1)$. Hence, we are allowed to use Taylor expansion of second order, that is,

$$f(x)=f(p)+\langle\nabla f(p), x-p\rangle+\frac{1}{2}\langle Hf(p)\cdot (x-p), x-p\rangle+r_p(x)$$

for every $x\in B(p, \delta_1)$, and

$$\lim_{x\to p} \frac{r_p(x)}{\|x-p\|^2}=0.$$ Above, $\langle\cdot, \cdot\rangle$ stands for the usual inner product in $\mathbb R^n$ and $\|\cdot\|$ stands for the induced norm.

Since $\nabla f(p)=0$, the previous expansion yields

$$f(x)-f(p)=\frac{1}{2}\langle Hf(p)\cdot (x-p), x-p\rangle+r_p(x).$$

Now, since

$$\lim_{x\to p} \frac{r_p(x)}{\|x-p\|^2}=0$$

there is $\delta_2>0$ such that:

$$\|x-p\|<\delta_2\implies \frac{|r_p(x)|}{\|x-p\|^2}<-\max_{\|v\|=1}\langle H f(p)\cdot v, v\rangle.$$ Notice that $-\langle Hf(p)\cdot v, v\rangle>0$ for every $v\in\mathbb R^n\setminus\{0\}$, so the above makes sense.

Also, since $\lim_{x\to p} \frac{\|x-p\|^2}{2}=0$ there is $\delta_3>0$ such that

$$\|x-p\|<\delta_3\implies \frac{\|x-p\|^2}{2}<2\Rightarrow \frac{\|x-p\|^2}{2}-1<1.$$

Then, whenever $\|x-p\|<\min\{\delta_1, \delta_2, \delta_3\}$, it follows:

$$\begin{align*} f(x)-f(p)&=-\frac{\|x-p\|^2}{2}\cdot \left(-\langle H f(p)\cdot \frac{x-p}{\|x-p\|}, \frac{x-p}{\|x-p\|}\rangle\right)+\frac{r_p(x)}{\|x-p\|^2}\\ &\leq -\frac{\|x-p\|^2}{2}\min_{\|v\|=1}\left(-\langle H f(p)\cdot v, v\rangle\right)+\frac{|r_p(x)|}{\|x-p\|^2}\\ &<-\frac{\|x-p\|^2}{2}(-\max_{\|v\|=1}\langle H f(p)\cdot v, v\rangle)+(-\max_{\|v\|=1}\langle Hf(p)\cdot v, v\rangle)\\ &<\max_{\|v\|=1}\langle Hf(p)\cdot v, v\rangle \left(\frac{\|x-p\|^2}{2}-1\right)\\ &<\max_{\|v\|=1}\langle Hf(p)\cdot v, v\rangle\\ &<0. \end{align*}$$ Is the above proof correct?

Thanks.

Ѕᴀᴀᴅ
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PtF
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  • I'm not sure what you're doing with $\delta_2$, and everything afterwards, because $\nabla f(p)=0$, so that inner product is $0$ for all $v$, hence the maximum is also $0$. You probably have the right idea, just not written out correctly. See my answer to Why do we need to determine the definiteness of the Hessian to decide what a critical point is? for the details (btw there I wrote $f$ to have domain $\Bbb{R}^n$, but clearly that's unnecessary; we can just work in some open ball around the point of interest, as you did with $\delta_1$). – peek-a-boo Jan 01 '22 at 21:37
  • oh you probably meant $Hf(p)$ instead of $\nabla f(p)$? That would make sense; you're using compactness of the unit sphere in $\Bbb{R}^n$ and the negative-definiteness assumption of the Hessian. But then again, because of the negative-definiteness assumption, you should be looking at $-\min \langle Hf(p)\cdot v,v\rangle$ no? Or simply $\max |\langle Hf(p)\cdot v,v\rangle| = \max \bigg(-\langle Hf(p)\cdot v,v\rangle\bigg)$? – peek-a-boo Jan 01 '22 at 21:43
  • As to $\delta_2$ there was really a typo, as well as below it. I really meant $Hf(p)$ not $\nabla f(p)$. And yes, I'm using the compactness of the sphere. Now, the main idea in the proof is to bound $r_a(x)/|x-p|^2$ correctly so that in the end we get something $<0$. The choise of $-\max_{|v|=1}\langle Hf(p)\cdot v, v\rangle$ did that. The choice of $-\min_{|v|=1}\langle Hf(p)\cdot v, v\rangle$ also provides a bound in the sphere, but I dind't check if it will lead to the correct signs. But at first sight it will also work. – PtF Jan 02 '22 at 01:17
  • As a matter of fact, $-\min_{|v|=1}\langle Hf(p)\cdot v, v\rangle$ make things a bit simpler, since I wouldn't have to use a property like $\min(-X)=-\max(X)$. – PtF Jan 02 '22 at 01:19

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