Exercises — Uniform continuity — difference from pointwise

$\forall \epsilon >0\exists \delta >0\forall x,y\in A:|x-y|<\delta ⟹|f(x)-f(y)|<\epsilon .$ The single difference: here $\exists \delta$ comes before the points $\forall x,y$, so $\delta =\delta (\epsilon )$ only. In pointwise continuity $\exists \delta$ comes after the point $\forall x_0$, so $\delta =\delta (\epsilon ,x_0)$ may peek at the point. ==One $\delta$ for everyone== is the whole story.

$|x^2-y^2|=|x+y||x-y|.$ The amplifying factor is $==|x+y|==$. On $\mathbb{R}$ it is unbounded, so the same input gap $|x-y|$ produces an ever-larger output gap far out. Look at figure below — the required input width shrinks as $x$ grows.

(a) T — take $\delta =\epsilon /L$. (b) F — $\sqrt{x}$ on $[0,\infty )$ is uniformly continuous but not Lipschitz (its slope blows up near $0$). (c) T — Heine–Cantor. (d) F — $(0,1)$ is not closed/compact; $1/x$ is a counterexample.

$|f(x)-f(y)|=|3x-3y|=3|x-y|$, so $f$ is Lipschitz with $L=3$. Uniform continuity: take $\delta =\epsilon /3$. For $\epsilon =0.03$, $\delta =0.01$. Check: $|x-y|<0.01\Rightarrow |f(x)-f(y)|=3|x-y|<0.03$. $\delta$ has no $x$ in it — that's uniformity.

For $x,y\in [0,5]$: $|x+y|\le 10$, so $|x^2-y^2|=|x+y||x-y|\le 10|x-y|.$ Thus $f$ is Lipschitz on $[0,5]$ with $L=10$; take $\delta =\epsilon /10$. For $\epsilon =0.1$, $\delta =0.01$. The bounded factor $|x+y|\le 10$ is exactly what $\mathbb{R}$ lacked — this is Heine–Cantor in action on a compact interval.

Given the inequality, choose $\delta ={\epsilon }^2$. Then $|x-y|<{\epsilon }^2⟹|\sqrt{x}-\sqrt{y}|\le \sqrt{|x-y|}<\sqrt{{\epsilon }^2}=\epsilon .$ For $\epsilon =0.02$, $\delta =0.0004$. Note $f^{\prime }(x)=\frac{1}{2\sqrt{x}}\to \infty$ near $0$, so $f$ is not Lipschitz — yet still uniformly continuous. Lipschitz is sufficient, not necessary.

Why the inequality holds: $(\sqrt{x}-\sqrt{y}{)}^2=x+y-2\sqrt{xy}\le x+y-2\min (x,y)=|x-y|$, since $\sqrt{xy}\ge \min (x,y)$.

Fix $\epsilon =1$. Given any $\delta \in (0,1)$, pick a small $x$ and its neighbour: $x=\delta ,y=\frac{\delta }{2}.$ Both lie in $(0,1)$, and $|x-y|=\delta /2<\delta$. But $|\frac{1}{x}-\frac{1}{y}|=|\frac{1}{\delta }-\frac{2}{\delta }|=\frac{1}{\delta }>1=\epsilon (\text{since }\delta <1).$ Close in input, far in output — the formal negation. The trouble is the boundary point $0$ (excluded), where the slope $-1/x^2$ blows up. See figure: two nearby inputs, huge output gap.

Continuity: composition of continuous maps on $(0,1)$ (where $x\ne 0$). Now break uniformity with $\epsilon =1$. Choose points where the sine hits opposite peaks: $x_n=\frac{1}{2\pi n+\frac{\pi }{2}},y_n=\frac{1}{2\pi n-\frac{\pi }{2}}.$ Then $\sin (1/x_n)=\sin (2\pi n+\frac{\pi }{2})=1$ and $\sin (1/y_n)=\sin (2\pi n-\frac{\pi }{2})=-1$, so $|f(x_n)-f(y_n)|=|1-(-1)|=2>\epsilon .$ Yet $|x_n-y_n|=\frac{\pi }{(2\pi n{)}^2-(\pi /2{)}^2}\to 0$ as $n\to \infty$. So for any $\delta$ pick $n$ large enough that $|x_n-y_n|<\delta$ — the output gap stays $2$. Not uniformly continuous.

Let $a_n=1/n\in (0,1)$. It is Cauchy (it converges to $0$, so its terms bunch together). Apply $f$: $f(a_n)=n$. The image $(1,2,3,\dots )$ is not Cauchy (terms differ by $1$ forever). A uniformly continuous function must send Cauchy sequences to Cauchy sequences, so $1/x$ cannot be uniformly continuous on $(0,1)$. This is a slick criterion: breaks a Cauchy sequence $\Rightarrow$ not uniformly continuous.

Fix $\epsilon >0$. By uniform continuity get ${\delta }_f$ for tolerance $\epsilon /2$ (on $f$) and ${\delta }_g$ for tolerance $\epsilon /2$ (on $g$). Let $\delta =\min ({\delta }_f,{\delta }_g)>0$ — it still depends on $\epsilon$ only. Then for $|x-y|<\delta$, $|(f+g)(x)-(f+g)(y)|\le |f(x)-f(y)|+|g(x)-g(y)|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .$ The triangle inequality + splitting $\epsilon$ in half is the workhorse. $\delta$ never saw $x$, so $f+g$ is uniformly continuous.

$f(x)=x$: $|x-y|\le 1\cdot |x-y|$, Lipschitz with $L=1$, so uniformly continuous with $\delta =\epsilon$. But $f\cdot f=x^2$, which the parent note (and figure s01) shows fails on $\mathbb{R}$: fix $\epsilon =1$; for any $\delta$ take $x=\frac{1}{\delta },y=\frac{1}{\delta }+\frac{\delta }{2}$, giving $|x^2-y^2|=(\frac{2}{\delta }+\frac{\delta }{2})\frac{\delta }{2}=1+\frac{{\delta }^2}{4}>1.$ So the product broke uniformity even though each factor kept it. (The rescue: on any bounded domain products of uniformly continuous functions stay uniformly continuous, because the functions are then bounded.)

Fix $\epsilon >0$. Uniform continuity of $g$ gives $\eta >0$ with $|u-v|<\eta \Rightarrow |g(u)-g(v)|<\epsilon$ (for $u,v\in B$). Feed $\eta$ as the target tolerance into $f$: uniform continuity of $f$ gives $\delta >0$ with $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\eta$. Chain them: $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\eta \Rightarrow |g(f(x))-g(f(y))|<\epsilon$. Both $\delta ,\eta$ depend on $\epsilon$ only — uniformity survives the composition.

($\Leftarrow$) If $\omega$ exists with $\omega (t)\to 0$, then given $\epsilon >0$ pick $\delta >0$ with $\omega (t)<\epsilon$ for all $t<\delta$ (possible since $\omega (t)\to 0$). Then $|x-y|<\delta \Rightarrow |f(x)-f(y)|\le \omega (|x-y|)<\epsilon$. Uniform. ($\Rightarrow$) Define $\omega (t)=\sup \{|f(x)-f(y)|:|x-y|\le t\}$. Uniform continuity says: for each $\epsilon$ there is $\delta$ making every such difference $<\epsilon$ when $t<\delta$, i.e. $\omega (t)\le \epsilon$ for $t<\delta$. Hence $\omega (t)\to 0$ as $t\to 0^{+}$, and by construction $|f(x)-f(y)|\le \omega (|x-y|)$. For $\sqrt{x}$: $\omega (t)=\sqrt{t}\to 0$. Solving $\sqrt{t}<\epsilon$ gives $t<{\epsilon }^2$, recovering $\delta ={\epsilon }^2$ from L2.3.

Take any sequence $x_n\to a^{+}$ inside $(a,b)$; it is Cauchy. By L3.3's principle (uniform continuity preserves Cauchyness), $(f(x_n))$ is Cauchy in $\mathbb{R}$, hence converges (real Cauchy sequences converge). To see the limit is the same for every such sequence, interleave two sequences $x_n\to a$, $x_n^{\prime }\to a$ into one sequence $x_1,x_1^{\prime },x_2,x_2^{\prime },\cdots \to a$: it is Cauchy, its image is Cauchy, so it converges to one value — forcing $\lim f(x_n)=\lim f(x_n^{\prime })$. That common value is ${\lim }_{x\to a^{+}}f(x)$. Defining $f(a)$ as this limit (and likewise $f(b)$) extends $f$ continuously to the closed, compact $[a,b]$, on which Heine–Cantor confirms uniform continuity.

Contrast $1/x$ on $(0,1)$: it is not uniformly continuous, and indeed ${\lim }_{x\to 0^{+}}1/x=+\infty$ does not exist finitely — no continuous extension. Consistent.

Answer: uniformly continuous iff $0<p\le 1$.

• $0<p\le 1$ (uniform): Here $x^p$ is concave, and one can show the sub-additive bound $|x^p-y^p|\le |x-y{|}^p$ for $x,y\ge 0$. Since $t\mapsto t^p$ is a valid modulus of continuity ($t^p\to 0$ as $t\to 0$), L5.1 gives uniform continuity; explicitly $\delta ={\epsilon }^{1/p}$. (For $p=1$, $x^p=x$ is Lipschitz.)
• $p>1$ (fails): slope $p{x}^{p-1}\to \infty$, and like $x^2$ we engineer a break. Fix $\epsilon =1$. For any $\delta >0$, take $x$ large and $y=x+\frac{\delta }{2}$. By the MVT, $x^p-y^p$... rather $y^p-x^p=p{c}^{p-1}\cdot \frac{\delta }{2}$ for some $c\in (x,y)$; as $x\to \infty$, $c^{p-1}\to \infty$, so the output gap $\to \infty >1$ while the input gap $=\delta /2<\delta$. No fixed $\delta$ works. (The parent's $x^2$ is the case $p=2$.)
• Boundary $p=1$: exactly Lipschitz, the crossover point — uniform.

So the threshold sits precisely at $p=1$: sub-linear/linear growth (bounded or tapering slope effect via the $t^p$ modulus) is safe; super-linear growth ($p>1$) lets the slope run away.