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With n $\geq$ 2 and f: $\mathbb{R}^n \rightarrow \mathbb{R}$

$$f(x) = (1+x_n)^3\sum\limits_{i=1}^nx^2_i+x^2_n$$ Show that x*=0 is the only stationary point and that x* is a strict local minimum of f, but not global minimum.

I tried to find the gradient of the function wiht n=2 and n=3 and then generalize to any n and then prove that x* is the only stationary point. But I was getting that for x = (0, -1), the gradient was zero, so x* not being the only stationary point.

Iago Victor
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    To prove that $0$ is a local minimum, notice that $1+x_n\geq 0$ in a neighbourhood of $0$. In order to prove it is not global, try keeping $x_1=\cdots=x_{n-1}=0$ and moving along the $x_n$ axis. – Julio Puerta Feb 23 '24 at 13:30

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We'll find the critical points by imposing $\partial_{x_j}f(\mathbf x^\ast)=\mathbf 0,\ j=\{1,2,...,n\}$: $$\partial_{x_j}f(\mathbf x)=3(1+x_n)^2\delta_{nj}\sum_{i=1}^nx_i^2+2(1+x_n)^3\underbrace{\sum_{i=1}^nx_i\delta_{ij}}_{x_j}+2x_n\delta_{nj}$$

For $j=n$ we have $$3(1+x_n)^2\sum_{i=1}^nx_i^2+2(1+x_n)^3x_n+2x_n=0$$ And this for $j\neq n$: $$2(1+x_n)^3x_j=0\implies x_j=0\lor x_n=-1$$

Substituing the solution $x_n=-1$ in the $j=n$ equation we'd encounter a contradiction as $-2\neq 0$. Thus, the right solution would be $x_j=0$. We now do the same as before knowing that $\sum_{i=1}^n x_i^2=x_n^2$ because $x_j=0$ for $j=\{1,...,n-1\}$: $$3(1+x_n)^2x_n^2+2(1+x_n)^3x_n+2x_n=0\implies [3(1+x_n)^2x_n+2(1+x_n)^3+2]x_n=0$$ The trivial solution would be $x_n=0$, this way getting what the OP affirmed, namely, $\mathbf x^\ast=\mathbf 0$. As for the other solution, I'll just omit it as I'm not sure what it means and I prefer to focus on the point of interest.

Now then we'll study the behaviour of that point using second derivatives and the hessian. Let's start by computing those derivatives: $$\partial_{x_kx_j}f(\mathbf x)=\left(1+3(1+x_n)\sum_{i=1}^nx_i^2\right)2\delta_{nj}\delta_{nk}+6(1+x_n)^2(\delta_{nk}x_j+\delta_{nj}x_k)+2(1+x_n)^3\delta_{jk}$$ Let's now discuss all combinations and substitute the $0$s:

For $j=k\neq n$ we have $$2(1+x_n)^3=2$$ For $j=k=n$: $$2\left(1+3(1+x_n)\sum_{i=1}^nx_i^2\right)+6(1+x_n)^2(x_j+x_k)+2(1+x_n)^3=2+2=4$$ For $j=n,k\neq n$: $$6(1+x_n)^2x_k=0$$ For $k=n, j\neq n$: $$6(1+x_n)^2x_j=0$$ At last, for $j\neq k\neq n$, it is trivially $0$.

Hence, the hessian at that point looks like this: $$\mathbf Hf(\mathbf 0)= \begin{pmatrix} 2 & 0 & \cdots & 0\\ 0 & 2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 4 \end{pmatrix}$$ All of its eigenvalues are either $2$ or $4$ ($\lambda_i>0$), and therefore, it corresponds to a local minimum. It isn't a global minimum because if you use Julio Puerta's hint ($x_{j\neq n}=0,x_n=t$), then, the function becomes $f(\mathbf 0,t)=[(1+t)^3+1]t^2$ and if you graph it you'll see $f(\mathbf 0, t<-2)<f(\mathbf x^\ast=\mathbf 0)=0$ this way tending to $-\infty$ so $\mathbf x^\ast=\mathbf 0$ wouldn't be a global minimum.

Joan S. Guillamet F.
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