Let $f:R \rightarrow R$ such that $$| f(x+y)-f(x)-f(y) |\le |x-y|,$$ for all $x, y \in R.$ How can I prove that $$\lim_{x\to 0}f(x) = 0\iff\lim_{x\to 0}xf(x) = 0? $$
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$\lim_{x\to 0} x=0$, $\lim_{x\to0} f(x)=0$ implies $\lim_{x\to 0}x\cdot f(x)=0$ – Michael Galuza Jul 16 '15 at 16:36
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You can't prove this. What if $f(x)=1$? – David C. Ullrich Jul 16 '15 at 16:38
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@DavidC.Ullrich, and you'll have $|1-1-1|\le |x-y|$, and what? – Michael Galuza Jul 16 '15 at 16:39
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From left to right the assertion is OK, but from right to left...it isn't. – Idris Addou Jul 16 '15 at 16:40
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then $1 \leq |x-y| \implies $ if you select two points less than 1 appart, the function $f$ cannot be used in the example. – FisherDisinformation Jul 16 '15 at 16:40
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@MichaelGaluza Say $f(x)=1$. True or False: (i) $\lim_{x\to0}f(x)=0$. (ii) $\lim_{x\to0}xf(x)=0$. (iii) $\lim_{x\to0}f(x)=0\iff\lim_{x\to0}xf(x)=0$. – David C. Ullrich Jul 16 '15 at 16:41
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1Sorry everyone, I misread the question. – David C. Ullrich Jul 16 '15 at 16:42
3 Answers
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$(\implies)$
Straightforward.
$(\impliedby)$
Edit: Following David C. Ullrich's comment, I turned the wlog assumption $f \geqslant 0$ into using $|f(x)|$ instead of $f(x)$ everywhere.
First note that substituting $x = y$ yields $|f(2x) - f(x) - f(x)| \leqslant 0$, so $f(2x) = 2f(x)$ for every $x \in \mathbb{R}$.
Suppose that $\displaystyle \lim_{x \to 0} f(x) \neq 0$. We will show that also $\displaystyle \lim_{x \to 0} x \cdot f(x) \neq 0$.
Exists such $\varepsilon > 0$ that there are arbitrarily small $x \neq 0$ with $|f(x)| \geqslant \varepsilon$. Let $\delta > 0$. There is such $x \neq 0$ that $|x| < \delta^2$ and $|f(x)| \geqslant \varepsilon$. There is a unique $n \in \mathbb{N}$ such that $\frac{\delta}{2} \leqslant |x| \cdot 2^n < \delta$ and then
$$|f(x \cdot 2^n)| = 2^n \cdot |f(x)| \geqslant \frac{\delta}{2 |x|} \cdot \varepsilon > \frac{\delta}{2 \delta^2} \cdot \varepsilon = \frac{\varepsilon}{2 \delta}.$$
Thus $|x \cdot 2^n| < \delta$ and
$$|x \cdot 2^n| \cdot |f(x \cdot 2^n)| \geqslant \frac{\delta}{2} \cdot \frac{\varepsilon}{2 \delta} = \frac{\varepsilon}{4}.$$
Adayah
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2Excellent. You might consider just writing $|f|$ everywhere instead of saying we can assume $f\ge0$; in the current version one (strictly speaking) needs to show that $|f|$ satisfies the hypothesis... – David C. Ullrich Jul 16 '15 at 17:09
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You're right, but I only need the facts that $|f|(2x) = 2 |f|(x)$ and $\displaystyle \lim_{x \to 0} |f|(x) \neq 0$, which are left as exercise. :D – Adayah Jul 16 '15 at 17:13
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Certainly - wasn't complaining, just suggesting that just writing $|f|$ everywhere might be easier to follow. – David C. Ullrich Jul 16 '15 at 17:21
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@David, I can't comment (not enough reputation), but $f(x)=1$ fails the first hypothesis at $x=y=1$.
Guillermo Mosse
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From right to left
If $x=y$, we have $$ |f(2x)-2f(x)|\le |x-x|\Longrightarrow f(2x)=2f(x)\Longrightarrow f(x)=ax, $$ where $a$ is a constant, and statement is obvious.
Michael Galuza
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This is assuming continuity, I think. No such assumption is made here. – Cameron Williams Jul 16 '15 at 16:52
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No, you can't. There are even functions with $f(x+y) = f(x) + f(y)$ yet discontinuous. – Adayah Jul 16 '15 at 17:31
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@Adayah It's possible that he can prove $f$ continuous assuming that $\lim xf(x)=0$. Those discontinuous functions you refer to don't satisfy this. (Any such function must be unbounded on every intervall.) – David C. Ullrich Jul 16 '15 at 17:34
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@Adayah, strictly speaking, I can prove that $g(x)=f(x+y)-f(x)$ is continous for every $y$ – Michael Galuza Jul 16 '15 at 17:36