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I have to prove that the function $f(x)=\frac{1}{x}$ on $(0,\infty)$ is not uniformly continuous (for the definition of uniform continuity see here).

I negated the definition of uniform continuity getting:

$$ \exists \epsilon>0 \ \text{s.t.} \ \forall \delta>0 \ \text{I can always find} \ x,y \in(0,\infty) \ \text{s.t.} \mid x-y\mid<\delta \ \text{but} \ \mid f(x)-f(y)\mid \geq \epsilon. $$

So I chose $\epsilon=1$. Set $y=\frac{x}{2}$. Then I have to find an $x$ such that this holds: $\mid x-y\mid = \frac{x}{2} <\delta$ & $\mid f(x)-f(y)\mid = \frac{1}{x} \geq 1$, which is equivalent to $ x<\min [2\delta,1] $.

Is this argument valid? My concern is if I'm allowed to choose $x$ depending on $\delta$. Thanks!

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2 Answers 2

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It is correct. You can choose x depending on $\delta$ because $x$ is quantified before $\delta$.

Alternatively, we have $$1/n - 1/(n+1) \to 0$$ but

$$f(1/n) -f(1/(n+1)) = n - (n+1) \to -1 $$

And by the equivalent sequence characterisation of uniform continuity, we conclude that $f$ cannot be uniformly continuous.

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Here a full answer (that i writte too to practice) but take into account that I am just a student so I hope it is correct. For me you are on the good way.

1 - First let recall the definition of a non uniformly continuous function.
It exists at least one $\epsilon_0>0$ such that for every $\delta>0$ that we can choose it will always exists at least $x$ and $y$ that verifies $|x-y|<\delta$ but $|f(x)-f(y)|>\epsilon_0$.
More formally: $\exists \epsilon_0>0 \; , \forall \delta>0 \; : \; \exists |x-y|< \delta \Rightarrow |f(x)-f(y)| \geq \epsilon_0$

2 - Now let pay attention to the following inequalities:
(1): for any $\delta>0$ given it exists $N=Max(1; \left \lceil 1/ \delta \right \rceil)$ s.t. $1/N < \delta$ . Moreover all $n \geq N$ verifies too this inequality (by assumption we are in $(0; \infty)$ ).
(2): $\forall n \in \mathbb{N} $ we have $|\frac{1}{1/n}-\frac{1}{n+1/n}|=|n-\frac{n}{n^2+1}|=|n(1-\frac{1}{n^2+1})|=|\frac{n^3}{n^2+1}| \geq 1/2$

3 - Now we can writte:
$\exists \epsilon_0 = \frac{1}{4}>0$ such that for any $\delta>0$ it will always exists ,with $n \geq N$ as define in (1), at least two points $x_n=n$ and $y_n=n+1/n$ that despite that verifying $|x_n-y_n|=|1/n| < \delta$ (by (1)) $ \Rightarrow|f(x_n)-f(y_n)|=|\frac{n^3}{n^2+1}|>1/4$.
More formally: $\exists \epsilon_0 = \frac{1}{4}>0 \; , \forall \delta>0 \; : \; \exists \; x_n = n, \; y_n=n+\frac{1}{n}$ with $n \geq max(1; \left \lceil 1/ \delta \right \rceil)$
By (1): $|x_n-y_n|< \delta$
By (2): $\Rightarrow |f(x_n)-f(y_n)|=|\frac{n^3}{n^2+1}| \geq \epsilon_0 = \frac{1}{4}$
Q.E.D.

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