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There is a sequence of functions $f_n(x)=\frac{x}{n}$. It converges pointwise on $R$ to $0$ (here $x$ can't be equal to $\infty$ because $\infty$ isn't in $R$): $$ \lim_{n\rightarrow \infty} f_n(x)=\lim_{n\rightarrow \infty}\frac{x}{n}=0 $$

For the sequence to be uniformly convergent we need to prove that $\|f_n-f\|\rightarrow 0$ as $n\rightarrow\infty$. We can always select $x>n$ because $n \neq \infty, n \in N, N\in R$ and $\infty$ isn't a part of $N$ or $R$: $$ \|f_n-f\|=\sup_{x\in R}\Big(|f_n(x)-f(x)|\Big)=\sup_{x\in R}\Big(\Big|\frac{x}{n}\Big|\Big) \geq\Big|\frac{n}{n}\Big| = 1 \neq 0 \quad n\rightarrow \infty $$

Could you please help me with the following questions:

1) Are my thoughts above correct (especially regarding $\sup_{x\in R}\Big(\Big|\frac{x}{n}\Big|\Big) \geq\Big|\frac{n}{n}\Big|$)?

2) Can we use another norm instead of $\sup$ and when we should do it?

3) How is it possible to use Cauchy condition for uniform convergence to prove that the sequence doesn't converge uniformly on $R$ and does converge uniformly on $[0,1]$?

$$ \|f_m - f_n\|<\epsilon \quad \forall m,n > N $$

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    How do you know the sup exists? – 2017-02-12
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    @Bernard Unfortunately, I can't answer this question. Can you give a hint? – 2017-02-12
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    Actually, you have uniform convergence on every compact subset of $\mathbf R$, but no uniform convergence on $\mathbf R$, which a very strong condition. Note that most theorems about properties of the limit of a sequence of functions are true for uniform convergence on every compacet. – 2017-02-12

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Your thoughts are correct. More specifically, $\lim_{x\to +\infty} |\frac xn|=+\infty$, so $\sup |f_n|=+\infty$, and that means $f_n$ cannot converge uniformly to $0$.

Now, you cannot use another norm, if you want to study uniform convergence.

There is no need to use Cauchy condition to study uniform convergence in $[0,1]$. Note that $\sup_{x\in [0,1]} |f_n(x)| = \frac{1}{n} \to 0$ as $n$ goes to infinity. If you want to use it anyway you can say that, given $n>m$ natural numbers,

$$\sup_{x\in[0,1]} |f_n(x)-f_m(x)|=\frac{1}{m}-\frac{1}{n}=\frac{n-m}{nm}\leq \frac{n}{nm}=\frac{1}{m}$$ So, for any $\varepsilon>0$, choose $N$ so that $\frac{1}{N}<\varepsilon$, and for $n,m >N$, you get $\sup_{x\in[0,1]} |f_n(x)-f_m(x)|<\varepsilon$.

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    Thank you for your reply! How to prove that the series doesn't converge uniformly on $R$ using the Cauchy condition? – 2017-02-12
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    Note that $f_n(x)-f_m(x)=x(\frac{1}{n}-\frac{1}{m})$, which is unbounded for every $n, m\in \mathbb N$. So, $\sup_{x\in \mathbb R}|f_n(x)-f_m(x)|=+\infty$ – 2017-02-12