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

Here’s my solution:

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

My question is whether I can let xx depend on nn like that, and whether what I did was valid.

Thanks!

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What’s your domain? R\Bbb{R}?

– Zachary Selk

2 days ago

Yes, forgot to mention that.

– ChrisWong

yesterday

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1 Answer

1

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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 nn sufficiently large,

1nsupx∈R|x|=supx∈R|xn−0|<1\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||x| is
unbounded.
This should answer your question as well: "xx can depend on nn" (but not the other way around).
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.
– ChrisWong
yesterday
No; note that xx is a dummy variable (I take the sup).
– parsiad
yesterday