Hey guys,
How do I prove that (1/nx) --> 0 pointwise but not uniformly on (0,1) as n tends to infinity.
thanks in advance, I appreciate your help.
On $\displaystyle (0,1)$ and any $\displaystyle \varepsilon>0$
$\displaystyle
\frac{1}{nx}<\varepsilon,\ \ \forall n>N=\left \lceil \frac{1}{\varepsilon x}\right \rceil
$
which proves pointwise convergence.
For uniform convergence for $\displaystyle \varepsilon>0$ we need to find an $\displaystyle N$ such that for all $\displaystyle x\in (0,1):$
$\displaystyle
\frac{1}{nx}<\varepsilon,\ \ \forall n>N$
But for $\displaystyle x<\frac{1}{n\varepsilon}$ where $\displaystyle n>N;\ \frac{1}{nx}>\varepsilon$ which contradicts what would have to be true for uniform convergence to hold.
CB
I think that's just a formatting code that's no longer used - nothing mathematical about it. Here's CB's reply without the "br" formatting codes:
-------------------------
On $\displaystyle (0,1)$ and any $\displaystyle \varepsilon>0$,
$\displaystyle \frac{1}{nx}<\varepsilon,\ \ \forall n>N=\left \lceil \frac{1}{\varepsilon x}\right \rceil$
which proves pointwise convergence.
For uniform convergence for $\displaystyle \varepsilon>0$ we need to find an N such that for all $\displaystyle x\in (0,1)$:
$\displaystyle \frac{1}{nx}<\varepsilon,\ \ \forall n>N$
But for $\displaystyle x<\frac{1}{n\varepsilon}$ where $\displaystyle n>N;\ \frac{1}{nx}>\varepsilon$ which contradicts what would have to be true for uniform convergence to hold.
CB
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By the way, why did you resurrect this 4-year-old thread?
- Hollywood