Continuity and Differentiation

f is a continuous tent function with f(0) = f(1) = 0, so its periodic extension is continuous; continuous periodic ⇒ uniformly continuous. g = f∘$x^2$ is continuous but its oscillations speed up as $x\to \infty ($the tents get compressed), so it is not uniformly continuous.

A continuous function on (0,1) is uniformly continuous iff it extends continuously to [0,1]. $e^{-1/x^2}$ → 0 as x → 0⁺, so it extends; the other three oscillate without a limit at 0.

(a) false: 1/x. (b) true: uniformly continuous on a bounded set ⇒ bounded. (c) true. (d) false: sin(1/x).

Uniform continuity on $\mathbb{R}$ forces at most linear growth. f(x) = x shows (a) fails; $\surd$|x| shows (b) fails.

(a) Hölder-½ with |$\sqrt{x}-\sqrt{y}$| $\le \surd$|x − y|. (b) oscillation speeds up: not uniformly continuous. (c) extends continuously to [0,1] (limit 0 at 0). (d) super-linear growth.

(b) sin(1/x) is bounded but not uniformly continuous. (d) is stronger than uniform continuity.

(1) Continuous on a compact interval, hence uniformly continuous. (2) Fails: |f(n + 1/n) − f(n)| $=2+1/n^2$ stays above 2 while the arguments get arbitrarily close. (3) |$\sqrt{x}-\sqrt{y}$| $\le \surd$|x − y|, so $\delta ={\epsilon }^2$ works. (4) Fails: the difference quotient $(\sqrt{x}-0)/(x-0)=1/\sqrt{x}$ blows up near 0, so no Lipschitz constant exists there — uniform continuity does not imply Lipschitz.

g(x) = f(x) − x is continuous with g(0) = f(0) ≥ 0 and g(1) = f(1) − 1 ≤ 0, so the intermediate value theorem gives g = 0 somewhere. The other three all fail: |x − 1/2| is not differentiable at 1/2 nor monotone, and $\sqrt{x}$ maps [0,1] to [0,1] continuously without being Lipschitz near 0.

F′ = f changes sign exactly once, so F is monotone on each side of one point: at most two real roots (and the constant of integration can realise 0, 1 or 2).

f′(0) = lim h sin(1/h) = 0, but f′(x) = 2x sin(1/x) − cos(1/x) has no limit at 0.

(a) Darboux. (b) MVT gives Lipschitz. (c) fails: $x^2\sin (1/x).(d)$ Darboux ⇒ f′ has constant sign ⇒ strictly monotone.

(b): f′/g′ = 1 + cos x has no limit although f/g → 1. (c): write as ln x /(1/x). (d): two applications give 1/2.

(1) is Darboux's theorem — derivatives take intermediate values whether or not they are continuous. (2) fails for $f(x)=x^2\sin (1/x),f(0)=0$, whose derivative exists everywhere but oscillates at 0. (3) is the mean value theorem: $f(b)-f(a)=f^{\prime }(\xi )(b-a)>0.(4)$ is Fermat's interior-extremum theorem, applicable since f is differentiable on all of $\mathbb{R}$.