# Prove convexity with hessian matrix

I've got a function

$$f(x)=(x_1-1)^2+\sum_{i=2}^n (x_i-x_{i-1})^2\quad \text{with x\in \mathbb{R}^n}$$

I want to show that this is (strictly) convex, so I thought the best approach might be to look at $$\nabla f^2(x)$$ because $$\nabla f^2(x)$$ positive semidefinite/pos. def. iif $$f(x)$$ is convex/strictly convex.

It's pretty clear that the Hessian must be like this: \begin{align*} H_f = \begin{pmatrix} 4 & -2 & 0 & 0 &... & 0&0 &0\\ -2 & 4 & -2 & 0 &... & 0&0&0\\ 0 & -2 & 4 & -2 &... & 0&0&0\\ ...&...&...&...&...&...&...&...\\ 0 & 0 & 0 & 0 &... & -2&4&-2\\ 0 & 0 & 0 & 0 &... & 0&-2&4\\ \end{pmatrix} \end{align*}

I can prove fast that this matrix is pos. semidef. (Using diagonal dominance). But I suspect that it's also pos. def.. My "slow" way would be to look at the minors and prove by induction.

Is there a fast way to prove it? Or should I use another method to prove convexity of $$f$$?

• This is a duplicate of math.stackexchange.com/questions/3903369/… though it takes some consideration to recognize that instead of $-2$ in one location, it has been shifted into two $-1$' in each case Commented Mar 26, 2022 at 16:53

Actually, for the Hessian matrix $$H_f$$ that you defined, the principal minors form a monotone increasing sequence.. If we denote them as $$\Delta_1$$, $$\Delta_2$$, $$\ldots$$, etc., then
$$0 < \Delta_1 < \Delta_2 < \Delta_3 < \cdots < \Delta_k$$ $$\cdots$$
$$\Delta_k > 0$$ for all values of $$k$$.
Hence, by Sylvester's test for positive definiteness (a sufficient condition), we conclude that the Hessian matrix $$H_f$$ is strictly positive definite. $$\blacksquare$$