I have a sequence of functions on the space of bounded functions $B(S)$, $f_n(x)= (nx+1)^{-1}\sin(nx)$ where $x$ lies in the set $S= [0,\pi]$. Here I am supposed to show that $f_n$ converges to $f$ pointwise and I know that $f=0$ since $\lim_{n\to \infty}(nx+1)^{-1}\sin(nx)=0$ for any $x\in[0,\pi].$ You can show that by the Squeeze Law. We're then meant to show that $f_n$ does not converge to $f$ with respect to the sup-metric: $d(f,g) = sup_{x∈S} {|f(x) − g(x)|}$, where $f, g ∈ B(S)$. That means showing that $d(f_n,f)$ does not converge to $0$ right? But no matter how I approach the question it appears to do so. Am I doing something wrong? Thanks in advance for any help.
Pointwise Convergence and Proving Non-Convergence in a Metric
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metric-spaces
1 Answers
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We have $|f_n(x)-f(x)|=\frac{|\sin(nx)|}{nx+1}$. For $x=1/n$ we get
$\frac{\sin(1)}{2}=|f_n(1/n)-f(1/n)| \le d(f_n,f)$ for all $n$
Thus: $d(f_n,f)$ does not converge to $0$.
You have not told us your approach, so nobody can tell you what you have done wrong !
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0I thought I may have done something wrong with the limit as $n$ goes to $infty$. I was simply trying to show $d(f_n,f)$ didn't approach $0$ from the definition. But thanks for the help! – 2017-02-07