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Let $f:(0,+\infty)\to\mathbb{R}$ be continuous and bounded. Let $\xi>0$. Show that there is a sequence $(x_n)$ in $(0,+\infty)$ with $x_n\to\infty$ s.t.

$$\lim_{n\to\infty}|f(x_n+\xi)-f(x_n)|=0.$$

I tried to prove by contradiction. Assume this was false, then for any sequence $(x_n)$ with $x_n\to\infty$ and for any $n\in\mathbb{N}$, there exists $\varepsilon>0$ and $n_0\geqslant n$ s.t.

$$|f(x_{n_0}+\xi)-f(x_{n_0})|\geqslant\varepsilon$$

I want to conclude that in this case, $f$ cannot be bounded. I was stuck here. Am I on the right track? Can we prove this directly? I mean, without proof by contradiction.

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We have to prove that for all $x_0$, $\inf_{x\geqslant x_0}|f(x+\xi)-f(x)|=0$. If we manage to do that, then that $x_1$ such that $|f(x_1+\xi)-f(x)|\lt 1$, $x_2\geqslant x_1+1$ such that $|f(x_2+\xi)-f(x_2)|\lt 1/2$ and more generally $x_{n+1}\geqslant x_n+n$ such that $|f(x_n+\xi)-f(x_n)|\leqslant n^{—1}$.

This infimum exists and is finite since $f$ is bounded. Fix $x_0$ and let $\alpha:=\inf_{x\geqslant x_0}|f(x+\xi)-f(x)|$. Assume that $\alpha$ is positive. Define $g(x):=f(x+\xi)-f(x)$: $g$ is continuous doesn't vanish on $[x_0,\infty)$, hence, considering $-f$ instead of $f$, we can assume that $g(x)\geqslant \alpha\gt 0$ for all $x\geqslant x_0$. Using this with $x:=x_0+j\xi$, we get that $$f(x_0+n\xi)-f(x_0)\geqslant n\alpha.$$ Since $f$ is bounded, we get $\alpha\leqslant \frac{2\sup|f|}n$, hence $\alpha=0$, a contradiction.

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  • $\begingroup$ This makes sense. Quite clear now,Thanks. $\endgroup$ – Kato yu Nov 19 '13 at 0:00
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Here's a proof, partly by contradiction, that separates the roles of the two conditions on $f$.

Let $F(x)=f(x+\xi)-f(x)$. Since $f$ is continuous, so is $F$. If $F$ takes the value $0$ infinitely often, we're done. Otherwise, without loss of generality (by reflecting across the $x$-axis and shifting the origin to the right, as necessary), we can assume that $F(x)\gt0$ for all $x$. Let $L=\liminf_{x\to\infty} F(x)$. If $L=0$, we're done. So we need only show that $L\gt0$ leads to a contradiction.

If $L\gt0$, then $F(x)\gt L/2$ for all $x$ greater than some $x_0$ (which we could also assume, without loss of generality, to be $0$, but let's not). But then

$$f(x_0+n\xi)=F(x_0+(n-1)\xi)+F(x_0+(n-2)\xi)+\cdots+F(x_0)+f(x)\gt nL/2+f(x_0)$$

implies $f$ is unbounded, which is the desired contradiction.

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