Prove/Disprove : Let $f:(0,1)→\mathbb{R}$ be Continuous. The condition $f(1/n)$$\rightarrow$$1/2$ and $f(1/n^2)$$\rightarrow$$1/4$ imply that $f $ is uniformly continuous.
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In fact, the condition implies $f$ is not uniformly continuous. – David Mitra Jan 22 '14 at 14:28
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any counterexample? @ David Mitra – Topology Jan 22 '14 at 14:29
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Actually, the two conditions can't hold simultaneously. – David Mitra Jan 22 '14 at 14:31
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1If $f$ were uniformly continuous, there would be a uniformly continuous extension $F\colon [0,1] \to \mathbb R$. – martini Jan 22 '14 at 14:32
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@DavidMitra Why not? $f$ is only defined on $(0,1)$. – Neal Jan 22 '14 at 14:33
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3@Neal $(1/n^2)$ is a subsequence of $(1/m)$. – David Mitra Jan 22 '14 at 14:33
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Oh ho ho ho, joke's on me. – Neal Jan 22 '14 at 14:34
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1Disregarding this, just argue directly: any nhood $(0,\delta)$ contains points $x$, $y$ with $|f(x)-f(y)|\ge 1/8$. – David Mitra Jan 22 '14 at 14:35
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I guess it would be more precise to say $\geq 1/8$. Regardless of the fact that this condition is absurd :p – TerranDrop Jan 22 '14 at 14:38
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@user48805 Of course, thanks. Corrected ... – David Mitra Jan 22 '14 at 14:39
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@drhab Of course. I'm just imagining the question was well-posed... – David Mitra Jan 22 '14 at 14:49
3 Answers
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The condition implies everything! This because it is not true.
($f\left(\frac{1}{n}\right)\rightarrow2$ implies: $f\left(\frac{1}{n^{2}}\right)\rightarrow2$ wich contradicts $f\left(\frac{1}{n^{2}}\right)\rightarrow4$)
$p\rightarrow q$ is true if $p$ is not true.
So indeed the condition implies that $f$ is uniformly continuous. There is no function that satisfies this condition and is not uniformly continuous. You cannot even find a function that satisfies this condition.
drhab
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there is no function that satisfy this condition. let $k \in (1,\infty)$ then $ f(\frac{1}{k}) \to \frac{1}{2}$ and $f(\frac{1}{k^2}) \to\frac{1}{4}$ .
now introduce some $m=\sqrt{k}$ what do you observe.
you get a $f(\frac{1}{k}) \to \frac{1}{4}$ this means that a value of $x$ has $2$ different values of function.
Suraj M S
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As pointed out in other answers, the conditions are self-contradictory; the sequence $f(1/n^2)$ is a subsequence of $f(1/n)$. If they have limits at all, they cannot have different ones.
It seems to me that the spirit of the question is to ask,
if $a_n,\ b_n\to 0$ but $\lim f(a_n)\neq \lim f(b_n)$, then can $f$ be uniformly continuous?
Now my hint is: $f$ cannot extend to a continuous function $\hat{f}:[0,1]\to\mathbb{R}$. (Can you see why?) Why does this imply $f$ must not be uniformly continuous?
Neal
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@Mathematics Hint: the value of $f$ at $0$ must not depend on the sequence used to approach $0$. – Neal Jan 22 '14 at 14:34
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1The conditions taken together also imply that $f$ can extend to a continuous function $\hat{f}:\left[0,1\right]\rightarrow\mathbb{R}$. In fact they imply any statement. – drhab Jan 22 '14 at 14:55