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As we learned in calculus, if a function $f:\mathbb{R} \rightarrow \mathbb{R}$ is twice differentiable, and has a local maximum at point $x_0$, then $f'(x_0) = 0$ and $f''(x_0) \leq 0$. In addition, it's not hard to show.

What I have trouble with is the higher dimensional analogue:

Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be twice differentiable, and has a local maximum at $x_0\in \mathbb{R}^n$, then $Df(x_0) = 0$ and $D^2f(x_0) \leq 0$ (i.e. the symmetric matrix $-D^2f(x_0)$ is positive semidefinite).

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up vote 2 down vote accepted

You can deduce this from Taylor's formula for the multi variable case. The below article is taken from my lecture notes. Actually the link for this article isn't working now. I shall add the link when it starts working.

Assume that $0 \in U$, $U$ is star shaped at $0$ and that we wish to find a 'Taylor expansion' of $f$ at $0$. Consider the one variable function $g(t) = f(tx)$. We first observe that this is defined for $t \in (-\epsilon, 1+\epsilon)$ for some sufficiently small $\epsilon >0$. For, since $U$ is open and $0 \in U$, there exists a $\epsilon_{1}>0$ such that $B(0,\epsilon_{1}) \subset U$. Hence $0 +tx \in B(0,\epsilon_{1})$ provided that $||tx|| = |t|\cdot ||x|| < \epsilon_{1}$, that is when $|t| < \frac{\epsilon_{1}}{||x||}$. A similar consideration will show that $x+tx= (1+t)x \in B(x,\epsilon_{2}) \subset U$ if $|t| <\frac{\epsilon_{2}}{||x||}$. If we choose $\epsilon = \min \{\epsilon_{1}/||x||, \epsilon_{2}/||x||\}$ then $tx \in U$ for $t \in (-\epsilon,1+\epsilon)$. Let us now show that $g$ is differentiable on this interval and compute it's derivative.

\begin{align*} g'(t) = \lim_{h \to 0} \frac{g(t+h)-g(t)}{h} &= \lim_{h \to 0} \frac{f((t+h)x -f(tx)}{h} \\ &=\lim_{h \to 0} \frac{f(tx+hx)-f(tx)}{h} \\ &=D_{x}f(tx) \equiv Df(tx)(x) \\ &= \sum\limits_{i=1}^{n}\frac{\partial{f}}{\partial{x_i}}(tx)x_{i} \end{align*}

In particluar $$g'(0)= \sum\limits_{i=1}^{n} \frac{\partial{f}}{\partial{x_i}} (0)x_{i}$$ Similarlay one can find that $$g''(t) = \sum\limits_{i,j=1}^{n} \frac{\partial^{2}{f}}{\partial{x_j}\partial{x_i}} (tx)x_{j}x_{i}$$

Now, we can apply Taylor's theorem of one variable calculus to $g$. We get $$g(t)=g(0)+g'(0)t + g''(0)t^{2} + R$$ where the remainder $R$ is such that $\lim_{t \to 0} R/t^2=0$. Taking $t=1$ in the above equation we get the Taylor's formula for $f$. $$ f(x)=f(0) + \sum\limits_{i=1}^{n}\frac{\partial{f}}{\partial{x_i}}(0)x_{i} + \sum\limits_{i,j=1}^{n} \frac{\partial^{2}{f}}{\partial{x_j}\partial{x_i}} (0)x_{j}x_{i} +R \qquad (\text{I})$$

From (I) it is easy to deduce the sufficient condition (in terms of second order partial derivatives) for the local maximum/minimum for $f$. If we assume for instance that $x=0$ is a point of local maximum, then $t=0$ is a point of local maximum for $g$. Hence $g''(0) \leq 0$. Coming back to $f$ this implies that $$ \sum\limits_{i,j=1}^{n} \frac{\partial^{2}{f}}{\partial{x_j}\partial{x_i}} (0)x_{j}x_{i} \leq 0$$ for all choices of $x$ in a neighborhood of $0$. This is same as saying that the matrix $$ D^{2}f(0) = \left(\begin{array}{cccccc} \frac{\partial^{2}{f}}{\partial{x_{1}^{2}}}(0) & \frac{\partial^{2}{f}}{\partial{x_1}\partial{x_2}}(0) & \cdot & \cdot & \cdot & \frac{\partial^{2}{f}}{\partial{x_1}\partial{x_n}}(0) \\\ & \cdot \\\ & \cdot \\\ & \cdot \\\ \frac{\partial^{2}f}{\partial{x_n}\partial{x_1}}(0) & \frac{\partial^{2}{f}}{\partial{x_n}\partial{x_2}}(0) & \cdot & \cdot & \cdot & \frac{\partial^{2}{f}}{\partial{x_{n}^{2}}}(0) \end{array}\right)$$

is negative semi-definite.

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By $f(x)$ do you mean $f(x_1, x_2, x_3 ...x_n)$? – user38268 Jul 3 '11 at 11:45
You used an $\epsilon-\delta$ argument above to show that $f(p+t(x-p))$ is defined. But since $U$ is star shaped w.r.t. the point p, for any $x$ in $U$ the line segment $p +(x-p)t$ lies in $U$. So $f(p +(x-p)t)$ is defined for $0 \leq t \leq 1$; can't you just say this directly? – user38268 Jul 3 '11 at 11:49

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