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

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 $$\delta\cdots$$ ?)

• You might find it nicer to write $(\sin(x))^2$ in terms of $\cos(2x)$. If you have access to the mean value theorem, then I would use it! Jul 25, 2021 at 6:46
• I would rather use periodicity and Heine theorem. Jul 25, 2021 at 6:51
• $\lvert \sin x+\sin y\rvert\lvert \sin x-\sin y\rvert<\epsilon$ is implied by $2\lvert \sin x-\sin y\rvert<\epsilon$ is implied by $\lvert \sin x-\sin y\rvert<\frac{\epsilon}{2}$. Use the uniform continuity of sine function to conclude the rest. Jul 25, 2021 at 6:54
• A more general result math.stackexchange.com/questions/775045/… Jul 25, 2021 at 7:07

$$|f'(x)|=|2\sin x \cos x| \leq 2$$. 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 $$|\sin x -\sin y|=|\int_x^{y} \cos t dt| \leq |\int_x^{y} 1 dt|=|x-y|$$.

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

• Please, write \sin(x) to display $\sin(x)$ i.e. add a backslash. Jul 25, 2021 at 7:18
• It's worth learning a few more MathJax tips. For big, complicated, or important equations, try using $$ instead of  to enclose your code. This renders your maths in display mode, e.g. $$\sin^2(x)=\frac{1}{2}-\frac{1}{2}\cos(2x).$$ renders like so:$$\sin^2(x)=\frac{1}{2}-\frac{1}{2}\cos(2x).I would also suggest having a look at this entry in the MathJax tutorial, for a much nicer way to render the really long equations (such as your first one). Jul 25, 2021 at 8:05
• I know mathjax and have been using it since 2018. I'll edit it . I wrote it from my mobile phone and I just copied the code from the op. Jul 25, 2021 at 8:42
• @ArghyadeepChatterjee Oops! I believe you. I didn't just make the assumption out of thin air; I checked most of your other answers and didn't find any display maths. But, I should have seen that you just copied the OP. Jul 25, 2021 at 10:06
• @TheoBendit I am mostly used to latex. And I use \displaystyle to format it in the same line . Otherwise I do not much like to bring an expression on the next line as I add a lot of parentheses to provide reasons or questions for a particular step. It has become a habit.... and I actually write anything as I would write in pen and paper. And being a little miserly in nature...I actually try to use as less space on the paper as possible and so I start writing every line from the left edge.lol. Jul 25, 2021 at 10:59

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 $$(4)$$.

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 $$(-\infty,\infty)$$.