I have been trying to prove the uniform convergence of sequence of functions defined by $f_n(x)=x^n$ on $[0,k]$ where $k<1$ by the epsilon definition of uniform convergence.
I have found the point-wise limit of the same..i.e. $f(x)=0$
Now let $\epsilon>0$ be given.
Then if the given sequence is uniformly convergent then we must have that there exists a natural number $m$ such that $|x^n-0|<\epsilon$ for all $n\geq m$ and for all $x$ in $[0,k]$, which on further calculations give that $$n>\frac{\log(1/\epsilon)}{\log(1/x)}$$
Now what natural number $m$ should I choose such that for all $n\geq m$ the above holds?
Please help!
Best Answer
we begin by observing that $$\forall x\in[0,k]\;\; |x^n|\le k^n$$
given $\epsilon>0$. we just need to find $ m\ge 0$ such that
$$n\ge m \implies k^n<\epsilon$$ or $$n\ge m \;\implies n\ln(k)<\ln(\epsilon)$$
which gives $$n\ge m \;\implies n>\frac{\ln(\epsilon)}{\ln(k)}$$
we know that $$\lfloor \frac{\ln(\epsilon)}{\ln(k)}\rfloor +1>\frac{\ln(\epsilon)}{\ln(k)}$$ so, we can take for example $$m=\lfloor \frac{\ln(\epsilon)}{\ln(k)}\rfloor +3$$