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Show that the bounded function $f(x) = \sin(x^2)$ is continuous, but not uniformly continuous, on the interval $[0,\infty)$.

The problem should be solved using the following definitions:

Definition of Continuous Function Let $X$ and $Y$ be metric spaces. We sat that a function $f:X\rightarrow Y$ is continuous if given any open set $V \subseteq Y$, its inverse image $f^{-1}(V)$ is an open subset of $X$.

Definition of Uniformly Continuous Function Let $(X,d_x)$ and $(Y,d_y)$ be metric spaces. If $E \subseteq X$, then we say that a function $f: X\rightarrow Y$ is uniformly continuous on $E$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x,y \in E$ we have $$d_x(x,y) < \delta \longrightarrow d_y(f(x),f(y)) < \epsilon$$

For the continuity part it is clear that $f^{-1}(a,b), (a,b) \subseteq (-1,1) = \{x:f(x) \in (a,b)\}$ which appears to be an open set however I doubt this is sufficient as a proof since it holds for any function $f$.

Not sure how to go about proving $f$ not uniformly continuous. Seems to be difficult to prove from the definition alone.

Bernard
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Pame
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  • Prove the lemma: if $f$ is uniformly continuous then $f$ has the following property: if $(x_n)$ and $y_n$ are sequences in the domain such that $d(x_n, y_n) \to 0$ as $n \to \infty$, then $d(f(x_n), f(y_n)) \to 0$ as well. Refute that property and you refute uniform continuity. – Henno Brandsma Sep 18 '20 at 13:29
  • Henno has a good idea, but let me maybe state it this way: If you are uniformly continuous, your choice of epsilon does not depend on your choice of points. The idea is to now show that there are points where $\epsilon$ will depend on them. – User203940 Sep 18 '20 at 13:52

2 Answers2

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First, write down the Inverse Negtive Proposition of Uniformly Continuous. Then we now how to prove a function is not uniformly continuous.

Then, let $a_n=\sqrt{2n\pi+\dfrac{\pi}{2}}$ and $b_n=\sqrt{2n\pi-\dfrac{\pi}{2}}$. It is easy to prove that when $n\rightarrow +\infty$, $a_n-b_n\rightarrow 0$, but $f(a_n)-f(b_n)=2$. Therefore we proved $f$ is not a Uniformly Continuous Function.

For continuity, I think your proof missed some key points. It's better to prove the solution subset is open by solving equations. (here we need to consider 3 consitions)

user823011
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Let's prove Henno's claim, and then in the comments you can try and fill in the blanks for how to finish things.

Henno claim: If $(x_n), (y_n)$ are sequences with $d(x_n, y_n) \rightarrow 0$ and $f$ is uniformly continuous, then $d(f(x_n), f(y_n)) \rightarrow 0$.

The goal is to show if $(x_n), (y_n)$ are sequences with $d(x_n, y_n) \rightarrow 0$ as $n \rightarrow \infty$, then $d(f(x_n), f(y_n)) \rightarrow 0$ as $n \rightarrow \infty$. We want to show that for all $\epsilon >0$ there is an $N$ so that for all $n \geq N$ we have $$ d(f(x_n), f(y_n)) < \epsilon.$$

By definition of uniform continuity, for all $\epsilon >0$ there is a $\delta >0$ so that $$d(x,y) < \delta \implies d(f(x), f(y)) < \epsilon.$$

Let's fix arbitrary $\epsilon > 0$. Then uniform continuity tells us that there is some $\delta > 0$ so the above property holds. Since $d(x_n, y_n) \rightarrow 0$, there exists an $N$ so that for all $n \geq N$, $d(x_n, y_n) < \delta$. The definition of uniform continuity then says that $d(f(x_n), f(y_n)) < \epsilon$, as desired.

How would one use this to prove something is not uniformly continuous? Find two sequences which get arbitrarily close together but for which the function applied to them does not get arbitrarily close together. What sequences are a good choice here?

User203940
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