$$ f(T) : \mathbb{R}^2 \to \mathbb{R}^2\\ f(T) = \begin{pmatrix} \sin(x)\sin(y) \\y\end{pmatrix} $$

I have to prove that f is a continuous function in $\mathbb{R}^2$ using $\epsilon$ and $\delta$

My try ended up in a problem

for all $\epsilon>0$ Exist $\delta>0$ so that $\lVert T-a\rVert < \delta =\epsilon$ for all $a\in \mathbb{R}^2$

to prove $\lVert f(T) -f(a)\rVert < \epsilon$

$$ \lVert f(T) -f(a)\rVert= \sqrt{\bigl[(\sin(x)\sin(y)-\sin(a_1)\sin(a_2)\bigr]^2+(y-a_2)^2} $$

how can I prove that $\bigl[(\sin(x)\sin(y)-\sin(a_1)\sin(a_2)\bigr]^2 <(x-a_1)^2$ ? or did i have a wrong start ?


1 Answer 1


$$|\sin x\sin y-\sin a_1\sin a_2|\\ =|\sin x\sin y-\sin a_1\sin y+\sin a_1\sin y-\sin a_1\sin a_2|\\ \le |\sin x-\sin a_1|\cdot|\sin y|+|\sin a_1|\cdot|\sin y-\sin a_2|\\ \le |x-a_1|+|y-a_2| $$


$$(\sin x\sin y-\sin a_1\sin a_2)^2\le (|x-a_1|+|y-a_2|)^2\le 2(x-a_1)^2+2(y-a_2)^2$$

I used $|\sin\alpha-\sin\beta|\le|\alpha-\beta|$ (simple application of Mean Value Theorem), and $(\alpha+\beta)^2\le 2\alpha^2+2\beta^2$

Then you can say:

$\lVert f(T)-f(a)\rVert\le\sqrt{2(x-a_1)^2+3(y-a_2)^2}\le\sqrt{3(x-a_1)^2+3(y-a_2)^2}=\sqrt{3}\lVert T-a\rVert$

so you may take $\delta=\frac{\epsilon}{\sqrt 3}$

  • $\begingroup$ thanks it was tricky to multiply with sin(a1)sin(y), but i can not here use Mean value theorem coz i don t know and i am not allowed to use it in order to proof that |sin a -sin b| <= |a -b| , is an other way to proof that ? $\endgroup$
    – Mohbenay
    Jan 28, 2017 at 19:43
  • $\begingroup$ ah i found it thank you :D $\endgroup$
    – Mohbenay
    Jan 28, 2017 at 19:47
  • $\begingroup$ You may convert the difference to product: $|\sin x - \sin y|=2\big|\sin\frac{x-y}{2}\big|\cdot\big|\cos\frac{x+y}{2}\big|\le 2\big|\sin\frac{x-y}{2}\big|$ and use only $|\sin x|\le |x|$, if it looks easier to you. $\endgroup$
    – Momo
    Jan 28, 2017 at 19:52
  • $\begingroup$ nah i will use this , let be $ c \in [a,b]$ ,$ and f(x) ⁼ sin(x) then |f(a)-f(b)| \leq |f'(c)||b-a|$ $and |f'(x)| = |cos(c)| \leq 1 \Rightarrow |f(a)-f(b)| \leq |b-a|$ $\endgroup$
    – Mohbenay
    Jan 28, 2017 at 20:02

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