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I want to show the sequence of real functions $(f_n)$ where $f_i(x)=\frac{x}{i}$ is not uniformly convergent, though it converges pointwise to $f=0$.

Here's my solution:

Let $\varepsilon=1$. Then given any $N$, we can let $n=N+1$ and $x=N+2$. Then $|\frac{x}{n}|=\left|\frac{N+2}{N+1}\right|>1$. Thus $|f_n-0|>\varepsilon$ and $(f_n)$ is not uniformly convergent.

My question is whether I can let $x$ depend on $n$ like that, and whether what I did was valid.

Thanks!

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  • $\begingroup$ What's your domain? $\Bbb{R}$? $\endgroup$ – user223391 Oct 21 '16 at 19:28
  • $\begingroup$ Yes, forgot to mention that. $\endgroup$ – ChrisWong Oct 22 '16 at 14:52
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Your solution is fine. Here's, the same idea, with (in my opinion) an easier presentation.

Suppose the sequence converges uniformly. Then, for all $n$ sufficiently large, $$\frac{1}{n}\sup_{x\in\mathbb{R}}\left|x\right|=\sup_{x\in\mathbb{R}}\left|\frac{x}{n}-0\right|<1$$ This is clearly a contradiction, since $|x|$ is unbounded.

This should answer your question as well: "$x$ can depend on $n$" (but not the other way around).

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  • $\begingroup$ In this proof, isn't n depending on x? It seems like you're saying given an x, we can find an n such that |x| is bounded, which is a contradiction. $\endgroup$ – ChrisWong Oct 22 '16 at 14:55
  • $\begingroup$ No; note that $x$ is a dummy variable (I take the sup). $\endgroup$ – parsiad Oct 22 '16 at 18:12

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