Uniform continuity

Uniform continuity requires one input tolerance to work at every point of the domain for a chosen output tolerance.

Choose the uniform continuity input tolerance corresponding to output tolerance one. A totally bounded domain is covered by finitely many balls of smaller radius centred in the domain. Every function value differs by less than one from the value at one of their finitely many centres. Taking the largest centre value bounds the whole function. Compactness or closedness of the domain is unnecessary.

A product of two uniformly continuous functions need not have uniform continuity on an unbounded domain. The functions $f(x)=g(x)=x$ give the counterexample $x^2$, since inputs $n$ and $n+1/n$ approach each other while their squared values remain separated. For bounded factors the estimate $|f(x)g(x)-f(y)g(y)|\le \Vert f{\Vert }_{\infty }|g(x)-g(y)|+\Vert g{\Vert }_{\infty }|f(x)-f(y)|$ restores uniform continuity.

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.

Every continuous $f:\mathbb{R}\to \mathbb{R}$ can be approximated with pointwise convergence by uniformly continuous functions. Set $c_n(x)=\max (-n,\min (x,n))$ and $f_n=f\circ c_n$. The map $c_n$ is Lipschitz continuous with constant one and $f{|}_{[-n,n]}$ is uniformly continuous by the Heine-Cantor theorem, so $f_n$ is uniformly continuous on the whole line. On every fixed bounded interval, $f_n=f$ eventually.

The modulus of continuity of a function $f$ isIt quantifies uniform continuity by recording the largest oscillation of $f$ across pairs of points at distance at most $\delta$.

Set $\mu (0)=0$ and extend $\mu (r)=1$ for $r\ge 1$. This positive, nondecreasing modulus of continuity is weaker than a linear Lipschitz continuity bound but still has ${\int }_0^{1}dr/\mu (r)=\infty$. At small distances it is equivalent to $r\log (1/r)$. A signed $r\log r$ cannot be an upper modulus at distances less than one. The planar vorticity velocity kernel maps $L^1\cap L^{\infty }$ into velocity fields with this modulus.

Every continuous map from a compact metric space to a metric space is uniformly continuous.

A uniform limit of uniformly continuous maps between metric spaces is uniformly continuous. Approximate the limit at both endpoints by one fixed member of the sequence and use its uniform continuity between them.

A function between metric spaces is uniformly continuous exactly when $d(x_n,y_n)\to 0$ implies $d(f(x_n),f(y_n))\to 0$ for every two sequences $(x_n)$ and $(y_n)$. Failure of uniform continuity supplies pairs with input distance below $1/n$ but output distance bounded below.