Under what conditions the square of a linear function is strictly convex?

Question

Consider the function $f: \mathbb R^n \to \mathbb R$ defined by $$ f(x) = (w_0 + x_1w_1 + x_2w_2 + ... + x_nw_n )^2 $$ , where $w_i \in \mathbb R$.

Is this function strictly convex? If no, is there any condition about $w_i$ under which we can be sure that this function is strictly convex?

@harfe if we write $(x^Tw +w_0)^2$ then the hessian matrix ($2ww^T$) is positive semi definite. But under what conditions is it positive definite? — alireza, May 02 '23 at 20:11
@harfe I think there is no way to make $ww^T$ positive definite. It is always positive-semi definite. Am I right? — alireza, May 02 '23 at 20:25

nowhere · Accepted Answer · 2023-05-02T22:56:34.890

It's exactly as you said in the comments. The Hessian is $2ww^T$, and it cannot be positive definite (unless $n = 1$).

Indeed, $\text{rank} \ ww^T = \text{rank} \ w = 1$. But positive definiteness requires $\text{rank} \ n$.

So, assuming you are interested in case $n > 1$, $f(x)$ is never strictly convex.

Remark: In general, Hessian being positive definite is sufficient for strict convexity, but not necessary. For example, $f(x) = x^4$ has zero second derivative at $x = 0$, but it's still strictly convex. However, for quadratic functions (which applies to your case), strict convexity is equivalent to Hessian being positive definite.

Under what conditions the square of a linear function is strictly convex?

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