Show that $f(x)=x\sin(2x)$ is uniformly continuous on $(0,1)$

Question

I don't know how easy this is to prove simply by definition of uniform continuity:

Let $I\subset R$ be a real interval.

A real function $f:I\to R$ is said to be uniformly continuous on $I$ if:

$$\text{for every }\epsilon > 0\text{ there exists }\delta>0\text{ such that the following property holds:}$$ $$\forall x,y∈I \text{ s.t. } |x−y|<\delta\text{ it happens that }|f(x)−f(y)|<\epsilon$$

I know it's possible to show for most functions algebraically by finding an appropriate δ, but simply can't find it. I can't even do so for regular continuity. I'm really struggling with how difficult the algebra is for this function.

I've seen other similar questions asked, and people suggested using the Heine-Cantor theorem, and some other theorems related to Metric Spaces. But I'm not working with metric spaces here (I think). This is a question for a Real Analysis module. We are learning Metric Spaces separately at the same time in a different module. Are there versions of these theorems (which would make this proof very easy) for working with functions like this without the domain and range being metric spaces?

Answer 1

This is a method that uses only Real Analysis:

$\lim_{x\to 0} x\sin x=0$ and $\lim_{x\to 1} x\sin x=\sin 1<\infty$

Since the two limits exist finitely and $f$ is continuous on $(0,1)$ so $f$ is uniformly continuous.

Answer 2

I'm going to use $d$ instead of $\delta$ because I'm on a cell phone. $$(x+d)\sin(2x+2d)-x\sin{2x}=d\sin(2x+2d)+x(\sin(2x+2d)-\sin(2x))$$ Note on this bounded interval we have $0<x<1$. Splitting it up like this should make the algebra a bit easier.

Answer 3

We have $f'(x)=\sin(2x)+2x \cos(2x)$, hence

$ |f'(x)| \le 1+2x \le 3 $ for $ x \in (0,1) $

If $x,y \in (0,1)$, the Mean Value Theorem gives therefore

$$|f(x)-f(y)| \le 3|x-y|$$

If $\epsilon $ is given, how you have to choose $ \delta$ ?

Answer 4

Here is a much simpler approach, if you learned this result already:

Close the interval.

$$f(x)=x \sin(2x)$$ is continuous on the closed interval $[0,1]$, therefore it is uniformly continuous. This implies that $f(x) $ is also uniformly continuous on $(0,1) \subset [0,1]$.

Answer 5

I am just a student so verify carefully my answer and I hope this is correct.

Prove by definition.

First of all let pay attention to the following inequalities:
1-$|\sin(x-y)|\leq|x-y|, \forall \; x-y \in \mathbb{R} $
2-$|\cos(x)| \leq 1$ and $|\sin(x)| \leq 1$
3-triangle inequality: $|x+y| \leq |x| + |y|$
4-now as it is given $x \in (0;1) \Rightarrow 0 \leq |x|\leq 1$

Now $\forall \epsilon>0, \; \exists \delta_{\epsilon}= \epsilon/4$ such that it is verifies for all $x, y \in (0;1)$ satisfying too $|x-y|<\delta_{\epsilon}$:
$|x\cdot\sin(2x)-y\cdot\sin(2y)|=|x\cdot\sin(2x)-y\cdot\sin(2x) +y\cdot \sin(2x)-y\cdot\sin(2y)|=|\sin(2x)(x-y)+y(\sin(2x)-\sin(2y))|$
By (3): $\leq|\sin(2x)||x-y|+|y||\sin(2x)-\sin(2y)|$
By (2): $ \leq |x-y|+|y|2|\cos(x+y)\sin(x-y)|$
By (1)+(2):$\leq |x-y|+2|y||x-y|=|x-y|(1+2|y|) $
By (4): $\leq |x-y|3$
And because (as written above): $|x-y|<\delta_{\epsilon}=\epsilon/4$ for any $\epsilon$ fixed. It comes: $|x-y|3 \leq 3\epsilon/4 \leq \epsilon$ Q.E.D