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The question is

$f_n(x) = ax^n + b \cos(x/n)$ is a sequence of functions where $f_n: [0,1] \to \mathbb{R}$. Determine for which $a,b \in \mathbb{R}$ values, $f_n$ is Cauchy w.r.t. the sup-norm in $C[0,1]$.

Since all $a,b,x$ and $n$ are variables, I can not imagine how the graph would be like for different values. I know I should show sufficient effort but in these type of questions I usually use graph of the function thus I really do not know where to start.

Can you give me some hints?

Thank you in advance.

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You must not think of this as a function of 4 variables, but as functions of $x$. You are supposed to look at fixed pairs $a, b$ and decide for such pairs whether the sequence with index $n$ converges uniformly or not. –  user20266 Dec 16 '11 at 11:55

1 Answer 1

up vote 3 down vote accepted

Some hints: $C[0,1]$ is complete; so, $\{f_n\}$ is Cauchy if and only if it converges to some $f\in C[0,1]$ (so, this $f$ must be continuous). Also, if $\{f_n\}$ converges in $ C[0,1]$, it must converge to its pointwise limit.

Now, compute the pointwise limit of your sequence. Using the above, you'll be able to conclude that the sequence is not Cauchy except for, perhaps, in one special case. But this case is easily dealt with.

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Thank you so much! –  marvinthemartian Dec 16 '11 at 12:31
    
I want to check something. The pointwise limit of $f_n$ is a+b when $x=1$ and b otherwise. Then it converges to a continuous function only if $a=0$. You said "the sequence is not Cauchy except in one special case", but isn't it Cauchy only if a=0? –  marvinthemartian Dec 16 '11 at 12:59
    
@marvinthemartian Yes, it is. –  David Mitra Dec 16 '11 at 13:09

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