# I Analysis of a general function with a specific argument

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1. Dec 26, 2016

### Quireno

Hello everybody,
I'm currently helping a friend on an assignment of his, but we are both stumbled on this exercise, I'm posting it here

We define a function $f$ which goes from $\mathbb{R}$ to $\mathbb{R}$ such that its argument maps as
$$x \mapsto 1+\frac{1}{2}\sin\left(\frac{1}{x}\right)$$
Show that
$$\forall x,y \in\mathbb{R}\quad |f(x)-f(y)|\leq \frac{1}{2}\left|\frac{1}{x}-\frac{1}{y}\right|$$
and deduce that
$$\forall x,y \in [1,\infty) \quad |f(x)-f(y)|\leq \frac{1}{2}\left|x-y\right|$$
That's it. The fact that it treats the limit of a general funtion with that argument is what is confusing, I tried to get a counterexample but haven't succeed. If anyone knows the first step on how to solve this or a hint it would be appreciated.

2. Dec 26, 2016

Suggestion: Try letting $u=1/x$ and $v=1/y$ to show the first relation. Then basically you need to show that $|sin(u)-sin(v)| \leq |u-v|$ . $\\$ editing... For the second part, If $|1/x-1/y|>...$, since $|1/x-1/y|=|(x-y)/(xy) |$, if x,y>1, clearly $|x-y|>|(x-y)/(xy) |$, so the second part follows if you can prove the first.

Last edited: Dec 26, 2016
3. Dec 26, 2016

### Quireno

Thank you, but my problem persists: I am dealing with a general function here. Let's say $u=x^{-1}$ and $v=y^{-1}$, now I have to prove that
$$\forall u,v \in\mathbb{R}\quad \left|f\left(1+\frac{\sin(u)}{2}\right)-f\left(1+\frac{\sin(v)}{2}\right)\right| \leq \frac{1}{2}\left|u-v\right|$$
but $f$ can be any function! What am I missing?

4. Dec 26, 2016

I thought the mapping gives you f(x). Maybe I misinterpreted the problem. But I think I may have, in fact, interpreted it correctly...Usually as I know it, the function is the mapping.

5. Dec 26, 2016

### Quireno

Do you mean that the function does not matter? Maybe you are right, but can you explain further?

6. Dec 26, 2016

My major in college was physics and not math, but usually when they say "there is a mapping" that defines the function. In this case $f(x)=1+(1/2)sin(1/x)$. That's how I interpreted it. Maybe some other members who are or were math majors can also give their inputs. @Ray Vickson @fresh_42

7. Dec 26, 2016

### Staff: Mentor

That's my interpretation, too. I don't get things like "any function" or "the function doesn't matter".
Unless otherwise stated, $f(x)=1+\frac{1}{2}\sin{\frac{1}{x}}$. However "goes from $\mathbb{R}$ to $\mathbb{R}$" is also mysterious, since $f(0)$ isn't defined. For the same reason the for all quantifier isn't correct.

I suppose this question should have been in the homework section, in which case I would demand some efforts by the OP to see where he actually got stuck and which mathematical environment may be used. As an $I$ labeled calculus question I would refer to the Taylor series of the sine function or the first differentiation.

8. Dec 26, 2016

### Quireno

Ok, I will try to do the exercise by treating the x as f(x). This could have been a misinterpretation of notation by myself. As for the expansion, I don't think it is neither necessary nor convenient.

9. Dec 26, 2016

### Staff: Mentor

But short. The limit definition of $\frac{d}{dx}\sin(\frac{1}{x})$ could also help. What do you mean by treating $x$ as $f(x)$? I think it should read:
$$f : \mathbb{R}-\{0\} \longrightarrow \mathbb{R}\, , \,x \longmapsto f(x) := 1+\frac{1}{2}\sin(\frac{1}{x})$$
or $f=\{(x,y)\,\vert \,y=1+\frac{1}{2}\sin(\frac{1}{x}) \wedge x\in \mathbb{R} \wedge x\neq 0\}$ if you like.

10. Dec 26, 2016

### Quireno

Nevermind me, I'm just confused.
Anyway, I have now the first part
$$\left|\left(1+\frac{\sin(u)}{2}\right)-\left(1+\frac{\sin(v)}{2}\right)\right|=\left|\frac{\sin(u)}{2}-\frac{\sin(v)}{2}\right|=\frac{|\sin(u)-\sin(v)|}{2}\leq \frac{1}{2}\left|u-v\right|$$
which can be proved by the mean value theorem
$$\cos x=\frac{\sin u-\sin v}{u-v}\qquad \Rightarrow \frac{\sin u-\sin v}{u-v}\leq1$$

11. Dec 26, 2016

### Staff: Mentor

Some absolute values added and it looks o.k. to me. I don't know what the substitution is good for though, but anyway. And the mean value theorem is a deluxe version of the first derivative. This is one of the reasons the homework template is for. It is not meant to annoy members. It shall tell us which formulas are allowed, and which are not. Perhaps you were looking for a geometrical proof. We cannot know.

12. Dec 26, 2016

### Quireno

Will use the homework forum next time. Thank you, I can handle the problem from here.

13. Dec 26, 2016

### Staff: Mentor

You might have to take some care of the situation if zero is between $x$ and $y$!