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Let a function $f: \mathbb R \rightarrow \mathbb R$ be such that $f(n)=n^2$ for $n \in \mathbb N$. Why $f$ is not uniformly continuous?

Thanks

Richard
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    what are you trying? – M.H Mar 18 '13 at 19:59
  • intuitively, uniform continuity means we can find one delta for every x value in our limit condition; but x^2 increases without bound over R, so how can we expect to have one delta? – Coffee_Table Mar 18 '13 at 20:13

3 Answers3

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Hint: Every uniformly continuous function $g$ is such that the sequence $(x_n)$ defined by $x_n=g(n+1)-g(n)$ is bounded.

Did
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Suppose it is. i.e. for every $\varepsilon>0$ there is some $\delta>0$ such that if $|x-y| \leq \delta$ then $|f(x)-f(y)| \leq \varepsilon$. Any interval $[n,n+1]$ can be broken into at most $\lceil \frac{1}{\delta} \rceil$ intervals with length$<\delta$ using some partition $x_0=n<x_1<\ldots<n+1=x_k$. Using the triangle inequality, $|f(n)-f(n+1)| \leq |f(n)-f(x_1)|+|f(x_1)-f(x_2)|+\ldots+|f(x_{k-1})-f(n+1)| \leq \lceil \frac{1}{\delta} \rceil \varepsilon$. But this can't hold for all $n$ because $|f(n)-f(n+1)|=1+2n$ grows larger than any bound.

user1337
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Recall that $f$ is uniformly continuous if and only if: for every sequences $(x_n)$ and $(y_n)$ such that $\lim_{n\to\infty}x_n-y_n=0$ we have $\lim_{n\to\infty}f(x_n)-f(y_n)=0$.

Now for counterexample: take $x_n=n+\frac{1}{n}$ and $y_n=n$.