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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.

The times of local maxima of a Brownian motion path are dense in the half-line almost surely, and every such local maximum is a strict local maximum. Density follows from nowhere monotonicity of Brownian motion and the extreme value theorem. Strictness follows by checking that maxima over separated rational compact time intervals are unequal; the increment across their gap supplies an independent nondegenerate normal distribution.

For fixed rational $0\le t_1<t_2<t_3<t_4$, writeBy independent increments, $Z$ is independent of $(V,W)$ and has normal distribution $N(0,t_3-t_2)$. Conditional on $(V,W)$, the displayed difference has a continuous probability density function; in particular it equals zero with probability zero. Taking the countable intersection over all such rational quadruples proves that, on one event of probability one, the maxima over any two separated rational closed intervals are distinct.

Suppose a local maximum at $t$ were not a strict local maximum. There is a neighbourhood in which every value is at most $B_t$, and, arbitrarily close to $t$, some other time $u$ strictly inside that neighbourhood has $B_u=B_t$. Choose separated rational closed intervals containing $t$ and $u$, both lying inside that neighbourhood. When one time is zero, use a first interval with left endpoint zero. Both interval maxima equal $B_t$, contrary to the preceding event.

Therefore every local maximum of Brownian motion is a strict local maximum almost surely, simultaneously over all times. This countable-interval argument avoids an invalid intersection of probability-one events over uncountably many candidate times.

A local minimum is strict when every other point of the domain of a function in some neighbourhood has a larger value. Strictness is local: two separated strict local minima can have the same value, as $x^4-2x^2$ at $\pm 1$ demonstrates. Negating the function gives a strict local maximum.