Hessian Test for Convexity and Concavity¶

Theorem¶

Let $C\subseteq\mathbb{R}^n$ be convex and open, and let $f\colon C\to\mathbb{R}$ be twice continuously differentiable.

Then $f$ is convex on $C$ if and only if $\boldsymbol{H}_f(\boldsymbol{x})$ is positive semidefinite for every $\boldsymbol{x}\in C$$C$.

The function $f$ is concave on $C$ if and only if $\boldsymbol{H}_f(\boldsymbol{x})$ is negative semidefinite for every $\boldsymbol{x}\in C$$C$.

If $\boldsymbol{H}_f(\boldsymbol{x})$ is positive definite for every $\boldsymbol{x}\in C$$C$, then $f$ is strictly convex. If $\boldsymbol{H}_f(\boldsymbol{x})$ is negative definite for every $\boldsymbol{x}\in C$$C$, then $f$ is strictly concave.

Interpretation¶

The Hessian test turns the geometric question of bending upward or downward into a matrix definiteness check at every point of the domain. It extends the one-variable criterion that a twice-differentiable function on an open interval is convex if and only if $f''\ge0$ there (convexity and concavity).

The bowl $x_1^2+x_2^2$ has Hessian $2\boldsymbol{I}$ at every point, positive definite, so it is strictly convex; the dome $-x_1^2-x_2^2$ has $-2\boldsymbol{I}$ and is strictly concave. The saddle $x_1^2-x_2^2$ has the indefinite Hessian $\operatorname{diag}(2,-2)$: along the $x_1$-axis the graph bends up, along the $x_2$-axis down, so it is neither.

The test reduces to one variable. On the segment from $\boldsymbol{x}$ to $\boldsymbol{y}$, put $\boldsymbol{d}=\boldsymbol{y}-\boldsymbol{x}$ and $g(t)=f(\boldsymbol{x}+t\boldsymbol{d})$$t\mathbf{d})$; the chain rule, applied twice, gives

and $f$ is convex exactly when every such $g$ is. So the condition must hold at every point of $C$, and $C$ must be convex, or segments leave the domain: $1/x^2$ has $f''>0$ on $\mathbb{R}\setminus\{0\}$$\{0\}$, but the segment from $-1$ to $1$ crosses the gap, so convexity there is not even defined. Definiteness at a general point is checked with Sylvester's criterion: for $-x^2-xy-y^2$ in the interactive example the leading minors are $-2$ and $3$ everywhere.

The strict statement only goes one way: $x^4+y^4$ is strictly convex, but its Hessian $\operatorname{diag}(12x^2,12y^2)$ is only positive semidefinite, not positive definite, on the coordinate axes (where $x=0$ or $y=0$$0$). The test pays off in optimization, because for a convex function every local minimum is a global minimum.

Key Results¶

• Positive semidefinite Hessian everywhere is equivalent to convexity under the stated smoothness and convex-domain assumptions.
• Negative semidefinite Hessian everywhere is equivalent to concavity under the same assumptions.
• Positive definiteness everywhere is sufficient for strict convexity, but not necessary in every case.

Examples¶

• For $f(x,y)=x^2+y^2$$y^2$, $\boldsymbol{H}_f=2\boldsymbol{I}$$2\mathbf{I}$, so $f$ is strictly convex.
• For $f(x,y)=-(x^2+y^2)$$y^2)$, $\boldsymbol{H}_f=-2\boldsymbol{I}$$-2\mathbf{I}$, so $f$ is strictly concave.

Prerequisites¶

• Hessian Matrix
• Definite Matrix
• Convex and Concave Functions on $\mathbb{R}^n$

Related Concepts¶

• Principal Minors and Sylvester's Criterion
• Global Maximum and Minimum on $\mathbb{R}^n$
• Unconstrained Extrema Conditions

Sources¶

• Math IV: Multivariate Calculus · Week 4, Lecture 1

Keywords: Hessian convexity test, Hessian concavity test, second derivative convexity test Symbols: element of, real numbers, subset

Last updated: 27 October 2026 11:52