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Check the function series $(f_{n})_{n \in \mathbb{N}}, f_{n}:]0,\infty[ \rightarrow \mathbb{R}$ for pointwise and uniform convergence and prove your results.

a) $f_{ n }\left( x \right) =\frac { n^{ 2 } }{ x } $

b) $f_{ n }\left( x \right) =\frac { x }{ n^{ 2 } } $

c) $f_{ n }\left( x \right) =\frac { n+x }{ nx } $

d) $ f_{n}(x)=\begin{cases} 1 & \text{for } x \in ]0,n]\\ 0& \text{for} \ x \in ]n,\infty [ \end{cases} $


I am not sure if correct but i found the pointwise convergence for a) to be $\infty$ and for b) it is $\lim _{ n\rightarrow \infty }{ x\frac { 1 }{ n^{ 2 } } } =0$ for c) $\lim _{ n\rightarrow \infty }{ \frac { n+x }{ nx } } =\lim _{ n\rightarrow \infty }{ (\frac { n }{ nx } +\frac { x }{ nx } ) } =\frac { 1 }{ x } $ whats with d) ? and how should i check for uniform convergence ?

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    $\begingroup$ d) Draw the graphs of the functions and see where would they converge to. For uniform convergence, find $\sup_x |f_n(x)-f(x)|$ and see whether these $\sup$s converge to $0$ as $n\to \infty$. $\endgroup$
    – Kumara
    Commented Jan 28, 2013 at 13:31

1 Answer 1

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  • a) $(f_n)_{n\in \mathbb{N}}$ does not converge pointwise nor uniform. $(\frac{n^2}{x})_{n\in \mathbb{N}}$ is not bounded and therefore it does not converge. Because it does not converge pointwise it won't converge uniform too.
  • b) $f_n(x)=\frac{x}{n^2}$ converges pointwise but not uniform.
  • c) $f_n(x)=\frac{n+x}{nx}$ converges uniform and therefore pointwise too.
  • d) Converges Pointwise against $1$ but not uniform
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