Local maximum ($f(x)\le f(x_{\ast })\text{near }{x}_{\ast }$)

A real-valued function $f:D\to \mathbb{R}$, with $D\subseteq {\mathbb{R}}^n$, has a local maximum at $x_{\ast }$ if some $\delta >0$ satisfiesOnly a neighbourhood within the domain of a function is compared. Equality is allowed; a strict local maximum requires $f(x)<f(x_{\ast })$ for every other point in that neighbourhood. A constant function therefore has a local maximum and a local minimum at every input, with neither strict. This is a local form of an extremum of a real-valued function.

For $f(x)=x^4-2x^2$, zero is a strict local maximum because $f(x)=x^2(x^2-2)<0=f(0)$ when $0<|x|<\sqrt{2}$. It is not a global maximum: values far away grow without bound. This distinction makes a local maximum useful for describing a nearby peak without claiming that it is the highest value anywhere.

Negating the function exchanges local maxima and local minima, including their strict versions. At a smooth interior local maximum, the critical point argument forces zero gradient; for a function twice continuously differentiable nearby, its Hessian matrix is negative semidefinite. These necessary conditions alone do not establish a local maximum.

A function has a strict local maximum at $t$ if some neighbourhood contains no other point $s$ with $f(s)\ge f(t)$. Unlike a non-strict local maximum, equality at another sufficiently nearby point is excluded. Use neighbourhoods relative to the domain, including when $t$ is an endpoint.