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A nonnegative function that is a uniformly continuous function on the interval $[0,\infty )$ and has a finite integral tends to zero. Otherwise uniform continuity gives disjoint intervals of a fixed width on which the function exceeds a fixed positive value, contradicting integrability. This supplies a useful decay argument for Lyapunov functions with merely continuous vector fields.

To make the fixed width explicit, suppose some $\eta >0$ occurs as a lower bound $f(t_n)\ge \eta$ along a sequence with $t_n\to \infty$. Choose $\delta >0$ from uniform continuity for output tolerance $\eta /2$. On every interval $(t_n-\delta /2,t_n+\delta /2)$ lying in $[0,\infty )$, the function is greater than $\eta /2$. Extract a subsequence whose centers are more than $\delta$ apart. Each disjoint interval contributes at least $\eta \delta /2$ to the integral, so the first $N$ contribute at least $N\eta \delta /2$. This contradicts finiteness of the integral. The same input width at every center is the essential step; continuity alone would allow progressively narrower peaks.

A standard form of the Kakutani solution of the Dirichlet problem uses a bounded domain $D\subset {\mathbb{R}}^d$, continuous Dirichlet boundary data $f\in C(\partial D)$, and regularity of every boundary point for the Dirichlet problem. The boundedness of $D$ makes $\partial D$ a compact set, so $f$ is a bounded function and a uniformly continuous function. A bounded domain with $C^2$ boundary is a sufficient geometric case; one must not omit boundary regularity for an arbitrary bounded domain.

One standard analytic characterization of a regular boundary point on a bounded domain is the existence of a positive harmonic barrier: for each $\xi \in \partial D$ there is a harmonic function $h_{\xi }\in C(\overline{D})$ with $h_{\xi }(\xi )=0$ and $h_{\xi }(x)>0$ for $x\in \overline{D}\setminus \{\xi \}$. This is the harmonic form of a barrier for the Dirichlet problem; the barrier characterization of regularity is a standard fact of potential theory.

Let $B$ be $d$-dimensional Brownian motion started at $x\in D$ and let ${\tau }_D=\inf \{t\ge 0:B_t\notin D\}$ be its Brownian exit time. Then the unique solution in $C(\overline{D})\cap C^2(D)$ isEquivalently, $u(x)={\int }_{\partial D}f(\xi ){\omega }_D(x,d\xi )$, where ${\omega }_D$ is harmonic measure, the exit probability distribution of Brownian motion. The harmonic function equation is $\Delta u=0$, with the Laplace operator convention $\Delta ={\sum }_j{\partial }_j^2$.

A uniformly continuous function has one input tolerance valid throughout its domain of a function. For $f:D\to \mathbb{R}$, with $D\subseteq \mathbb{R}$, uniform continuity meansChoose the output tolerance $\epsilon$ first, then a positive $\delta$ valid for every pair of function inputs. It may depend on $\epsilon$, the function and its domain of a function, but not on the pair’s location. The absolute values measure input and output distance.^[ref]

Compare the quantifier order with the epsilon-delta definition of continuity:Both lines end with $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. In the first, $\delta$ can depend on the point $a$; in the second, it is chosen before $a$. Thus uniform continuity implies continuity at every point. Each requested $\epsilon$ gets its own shared tolerance.

For an exact example, take the polynomial function $f(x)=x^2$ on the closed interval $D=[0,2]$. For every pair of function inputs in that interval, polynomial factorization givesConsequently $\delta =\epsilon /4$ works everywhere on this domain of a function. With $\epsilon =0.4$, the common tolerance is $\delta =0.1$. Near the right endpoint, for example, $x=1.92$ and $y=2$ have input distance $0.08$ and output distance $4-1.92^2=0.3136<0.4$. The bound is sufficient; it does not claim the largest admissible input window at each point.

On the whole real line, the same polynomial function is a continuous function but fails uniform continuity. Take the sequences $x_n=n$ and $y_n=n+1/n$, for positive integers $n$. ThenFor any proposed $\delta >0$, choose an integer $n>1/\delta$. The inputs are closer than $\delta$, yet their outputs violate the fixed tolerance $\epsilon =1$. The sequences need not themselves have a finite limit: their mutual distance is what tends to zero. This is the obstruction captured by the sequential criterion for uniform continuity. Restricting the domain of a function to $[0,2]$ removes these escaping pairs.

A uniformly continuous function need not be a bounded function. The identity function $f(x)=x$ on the real line has $|f(x)-f(y)|=|x-y|$, so $\delta =\epsilon$ always works, although its image of a function is the unbounded real line. Uniform continuity controls changes between nearby inputs, not the overall size of outputs. Likewise, an unbounded domain of a function does not by itself prevent uniform continuity.

The square root $g(x)=\sqrt{x}$ supplies a different common tolerance on the interval $[0,\infty )$. Assuming $x\ge y\ge 0$, put $u=\sqrt{x}$ and $v=\sqrt{y}$. Since $0\le u-v\le u+v$,Interchanging the inputs when necessary provesThus $\delta ={\epsilon }^2$ guarantees uniform continuity on the entire domain of a function, including zero. For $\epsilon =0.4$, any input distance below $0.16$ suffices. The pairs $(0,0.09)$ and $(4,4.09)$ both have input distance $0.09$; their output distances are respectively $0.3$ and $0.09/(\sqrt{4.09}+2)<0.0225$. The common guarantee accommodates quite different local sensitivities.

This Hölder continuity bound allows a curved tolerance rule. It is weaker than a Lipschitz bound: at zero the output change divided by the input change isNo finite Lipschitz constant works near zero, even though uniform continuity does. A Lipschitz function does have uniform continuity, choosing $\delta =\epsilon /L$ when its Lipschitz bound $L$ is positive; when $L=0$ it is a constant function.^[ref]

For general metric spaces, replace the absolute values by the respective distance functions. The Heine-Cantor theorem gives an existence guarantee: every continuous function on a compact metric space is uniformly continuous. In particular, continuity on a finite closed interval provides a common tolerance, even when no simple formula for it is known.^[ref]

For a short proof outline, fix $\epsilon >0$. Continuity gives each point $a$ an open ball of radius $r_a>0$ within which outputs differ from $f(a)$ by less than $\epsilon /2$. The open balls of half these radii still cover the domain of a function. Compactness supplies a finite subcover. Take $\delta$ to be the smallest half-radius in this finite set. Two inputs closer than $\delta$ lie together in one of the original open balls, so the triangle inequality bounds their output distance by $\epsilon$. The beautiful step is that the compact set turns local choices into finitely many choices, whose minimum stays positive.

A shared tolerance also makes uniformly continuous functions preserve Cauchy sequences: after choosing $\delta$ for $\epsilon$, all sufficiently late input pairs are closer than $\delta$, so all corresponding output pairs are closer than $\epsilon$. Unlike a convergent sequence argument, this does not require an input limit inside the domain of a function. It is why uniform continuity is useful for extending a function toward missing endpoints and for the decay argument below.

If a uniformly continuous function $u:{\mathbb{R}}^d\to \mathbb{R}$ belongs to Lp space for some $1\le p<\infty$, then $u(x)\to 0$ as $|x|\to \infty$. Otherwise choose points escaping to infinity with $|u(x_j)|\ge 2\theta >0$. Uniform continuity gives one radius $\rho >0$ on which $|u|\ge \theta$ around every $x_j$. Extract disjoint such balls. Their contributions to $\int |u{|}^p$ are each at least ${\theta }^p|B_{\rho }|$, a contradiction. The finite-$p$ condition is essential: the constant function one lies in $W^{1,\infty }$ and does not vanish at infinity.