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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 $\delta\cdots$ ?)

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    $\begingroup$ 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! $\endgroup$ Jul 25, 2021 at 6:46
  • $\begingroup$ I would rather use periodicity and Heine theorem. $\endgroup$
    – nicomezi
    Jul 25, 2021 at 6:51
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    $\begingroup$ $ \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. $\endgroup$ Jul 25, 2021 at 6:54
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    $\begingroup$ A more general result math.stackexchange.com/questions/775045/… $\endgroup$
    – Robert Z
    Jul 25, 2021 at 7:07

4 Answers 4

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

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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

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  • $\begingroup$ Please, write \sin(x) to display $\sin(x)$ i.e. add a backslash. $\endgroup$
    – VIVID
    Jul 25, 2021 at 7:18
  • $\begingroup$ 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). $\endgroup$ Jul 25, 2021 at 8:05
  • $\begingroup$ 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. $\endgroup$ Jul 25, 2021 at 8:42
  • $\begingroup$ @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. $\endgroup$ Jul 25, 2021 at 10:06
  • $\begingroup$ @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. $\endgroup$ Jul 25, 2021 at 10:59
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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 $(4)$.

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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 $(-\infty,\infty)$.

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