example of tests for local extrema in Lagrange multiplier method

Let $n\in {\mathbb{N}}^{+}$ and $c\in ℝ$. We want to find the local extrema of the function

subject to the condition $g=0$, where

$$g:{ℝ}^n\to ℝ,x\mapsto \sum _{1\le i\le n}{x}_i-c.$$

The first and second order partial derivatives are for all $i,j\in \{1,\mathrm{\dots },n\}$

$${\partial }_{i}f(x)=\sum _{k\ne i}{x}_k,{\partial }_{i}g(x)=1,$$

$${\partial }_i{\partial }_{j}f(x)=1-{\delta }_{i,j},{\partial }_i{\partial }_{j}g(x)=0,$$

where ${\delta }_{i,j}$ is the Kroenecker-delta. Thus the necessary condition $f^{\prime }(x)=\lambda g^{\prime }(x)$ together with $g(x)=0$ gives the system of equations

$$\sum _{j\ne i}{x}_j=\lambda ,i\in \{1,\mathrm{\dots },n\},$$

$$\sum _{1\le j\le n}{x}_j=c.$$

By summing the first $n$ equations and then substituting in the last we get

$$(n-1)c=n\lambda ,$$

$$x_i=\sum _{1\le j\le n}{x}_j-\sum _{j\ne i}{x}_j=c-\lambda =\frac{c}{n},i\in \{1,\mathrm{\dots },n\}.$$

Thus there is only one point, where local extremum is possible. We apply the test in the parent entry to the matrix

$$D^2(f-\lambda g)(x)={[1-{\delta }_{i,j}]}_{i,j=1}^n=nP-I,$$

where $P$ is the matrix containing $1/n$ in all entries, and $I$ is the identity matrix. $P$ is a rank one projection. Therefore the second derivative has spectrum $\sigma (nP-I)=\{n-1,-1\}$, where $-1$ has multiplicity $n-1$, and $n-1$ has multiplicity $1$. Thus the second derivative of $f-\lambda g$ is indefinit, so it has no local extrema. However the nullspace of $g^{\prime }(x)$ is precisely the nullspace of $P$, thus the second derivative is strictly negative on the tangent space $T_x(M)$, so the vector $(c/n,\mathrm{\dots },c/n)$ is a local maximum of $f$ subject to $g=0$.