$f(x)=(\sin(x))^2$is uniformly continuous

Question

Q:
Given a function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x)=(\sin(x))^2$,prove $f(x)$is uniformly continuous.

I tried to go by definition:$|x-y|<\delta\Rightarrow|f(x)-f(y)|<\epsilon$
I transformed as followed
$|f(x)-f(y)|=|(\sin(x))^2-(\sin(y))^2|=|\sin(x)+\sin(y)||\sin(x)-\sin(y)|\leq1\cdots?$
I don't find how do I transform the equation.(When and How do I connect $\epsilon$ and $\delta\cdots$ ?)

Answer 1

I'll just write till your last step...$|f(x)-f(y)|=|(\sin(x))^2-(\sin(y))^2|=|\sin(x)+\sin(y)||\sin(x)-\sin(y)|\leq 2|\sin(x)-\sin(y)|$

Now $\sin(x)$ is a differentiable function.....hence you can write $\sin(x)-\sin(y)\leq (x-y)\cos(\theta)\leq x-y$ where $x\leq\theta\leq y$ or $y\leq\theta\leq x$ depending on if $x\leq y$ or $y\leq x$ So $|f(x)-f(y)|=|(\sin(x))^2-(\sin(y))^2|=|\sin(x)+\sin(y)||\sin(x)-\sin(y)|\leq 2|\sin(x)-\sin(y)|\leq 2|x-y|$. Now if for chosen $\epsilon>0$ and $\delta=\frac{\epsilon}{2}$ you have your required condition. What I used here is the Lipschitz criteria. Every continuously differentiable function satisfies lipschitz and hence is uniformly continuous

Answer 2

$|f'(x)|=|2\sin x \cos x| \leq 2$. By MVT $|f(x)-f(y)| \leq 2|x-y|$.

Alternatively, $|f(x)-f(y)|=|\sin x+\sin y||\sin x-\sin y|\leq (1+1)|\sin x-\sin y|$ and $|\sin x -\sin y|=|\int_x^{y} \cos t dt| \leq |\int_x^{y} 1 dt|=|x-y|$.

Answer 3

A more general result:

The product of bounded and uniformly continuous functions is again bounded and uniformly continuous.

Proof: Exercise. $\square$

Using this result, take $f=g=\sin:\mathbb{R}\to\mathbb{R}$.

To slightly mirror the proof of the general result, if we know a priori that $\sin:\mathbb{R}\to\mathbb{R}$ is uniformly continuous on $\mathbb{R}$ and bounded by $1$, then for any $\varepsilon>0$, there is a $\delta>0$ such that if $x,y\in\mathbb{R}$ with $|x-y|<\delta$, then $|\sin(x)-\sin(y)|<\varepsilon/2$.

Now, for any $x,y\in\mathbb{R}$ with $|x-y|<\delta$, we have $$\begin{align} |\sin(x)^2-\sin^2(y)| &=|\sin^2(x)-\sin(x)\sin(y)+\sin(x)\sin(y)-\sin^2(y)|\tag{1}\\ &\leq|\sin^2(x)-\sin(x)\sin(y)|+|\sin(x)\sin(y)-\sin^2(y)|\tag{2}\\ &=|\sin(x)||\sin(x)-\sin(y)|+|\sin(y)||\sin(x)-\sin(y)|\tag{3}\\ &\leq 2|\sin(x)-\sin(y)|\tag{4}\\&<2\cdot\frac{\varepsilon}{2}=\varepsilon \end{align}$$ where we used the triangle inequality between lines $(1)$ and $(2)$ and the fact that $|\sin(x)|\leq 1$ for all $x\in\mathbb{R}$ between lines $(3)$ and $(4)$.

Answer 4

If you have access to the following theorem, then the proof is trivial:

If $f$ is continuous on $[a,b]$, then it is uniformly continuous on $[a,b]$.

Since $\sin^2$ is the product of two differentiable functions, it is differentiable, and so it is continuous. In particular, $\sin^2$ is continuous on $[0,2\pi]$. Hence, it is uniformly continuous on $[0,2\pi$]. By periodicity of $\sin^2$, it is therefore uniformly continuous on $(-\infty,\infty)$.