MHB Problem of the Week #112 - July 21st, 2014

[Chris L T521]

Here's this week's problem!

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Problem: For each index $n$, define $f_n(x)=\alpha x^n+\beta\cos(x/n)$ for $0\leq x\leq 1$. For what values of the parameters $\alpha$ and $\beta$ is the sequence $\{f_n\}$ a Cauchy sequence in the metric space $C[0,1]$?
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EDIT: The norm on $C[a,b]$ is the maximum norm (in essence, the supremum norm) $\|f\|_{\max} = \max\{|f(x)|\mid x\in[a,b]\}$ and hence the metric is $d(f,g) = \|f-g\|_{\max}$. (Many thanks to Euge for inquiring about this since it wasn't specified in the original problem statement).

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Euge and Opalg. Ackbach gets honorable mention for getting the conditions half correct. You can find Opalg's solution below.

[sp]The space $C[0,1]$ is complete in the supremum norm. So the sequence will be Cauchy if and only if it converges in that norm. The sequence $\{\cos(x/n)\}$ converges uniformly to the constant function $1$. Therefore $\{\beta\cos(x/n)\}$ will be Cauchy for all $\beta$. The sum of two Cauchy sequences is Cauchy, so it follows that $\{f_n(x)\}$ will be Cauchy if and only if $\{\alpha x^n\}$ is Cauchy.

The sequence $\{x^n\}$ converges pointwise to $0$ if $x<1$, and to $1$ when $x=1.$ The uniform limit, if it existed, would have to be the same as the pointwise limit. But since that is discontinuous it follows that $\{x^n\}$ does not converge in $C[0,1]$ and is therefore not Cauchy. So the only value of $\alpha$ for which $\{\alpha x^n\}$ is Cauchy is $\alpha=0$.

Conclusion: the condition on $\alpha$ and $\beta$ is that $\alpha=0$; there is no restriction on $\beta$.[/sp]