Pointwise and uniform convergence

Nov 2008
76
2
I got stuck on two problems that involves finding the pointwisse limits of piecewise functions. Hope someone can give me a hand.
1) Let \(\displaystyle \{a_n\}\) be a sequence that contains each rational in [0,1] precisely once. For each \(\displaystyle n\in N\), define
\(\displaystyle f_n(x)= 0\) if x is irrational
\(\displaystyle f_n(a_i)=0\) if \(\displaystyle i > n\)
\(\displaystyle f_n(a_i)=1\) if \(\displaystyle i \leq n\)
Prove that \(\displaystyle \{f_n\}\) converges pointwise on [0,1] to a function that is not R-integrable.
My attempt:
I guess that this sequence of functions converges to the rational characteristic function, but I can't really show \(\displaystyle lim f_n=f\) as n approaches infinity.
I tried this again and I got this far: \(\displaystyle f(x)=0\) if x is irrational and \(\displaystyle f(x)=1\) if x is rational.
Consider: absolute value of \(\displaystyle f_n(x)-f(x)\)=\(\displaystyle 0-0=0< \epsilon\) if x is irrational. Choose \(\displaystyle \epsilon=1/2\)
If x is rational, \(\displaystyle f_n(x)-f(x)=1-1=0< \epsilon\). I think f(x)=1 because eventually \(\displaystyle i \leq n\) as n approaches infinity.

2) For \(\displaystyle n \in N\) define \(\displaystyle f_n: [0,1] \rightarrow R\) by \(\displaystyle f_n(x)=1/x\) if \(\displaystyle x \in [1/n,1]\) and \(\displaystyle f_n(x)=n^2x\) if \(\displaystyle x \in [0,1/n)\). Prove that \(\displaystyle \{f_n\}\) does not converge uniformly on [0,1] but converge uniformly on [M,1] where 0<M<1.

I don't know what the pointwise limit of \(\displaystyle f_n\), so I can't go any anywhere. (Crying) It would be really helpful if someone helps me how to find the pointwise limit
 
Last edited:
Jul 2009
555
298
Zürich
I got stuck on two problems that involves finding the pointwisse limits of piecewise functions. Hope someone can give me a hand.
1) Let \(\displaystyle \{a_n\}\) be a sequence that contains each rational in [0,1] precisely once. For each \(\displaystyle n\in N\), define
\(\displaystyle f_n(x)= 0\) if x is irrational
\(\displaystyle f_n(a_i)=0\) if \(\displaystyle i > n\)
\(\displaystyle f_n(a_i)=1\) if \(\displaystyle i \leq n\)
Prove that \(\displaystyle \{f_n\}\) converges pointwise on [0,1] to a function that is not R-integrable.
My attempt:
I guess that this sequence of functions converges to the rational characteristic function, but I can't really show \(\displaystyle lim f_n=f\) as n approaches infinity.
I tried this again and I got this far: \(\displaystyle f(x)=0\) if x is irrational and \(\displaystyle f(x)=1\) if x is rational.
Consider: absolute value of \(\displaystyle f_n(x)-f(x)\)=\(\displaystyle 0-0=0< \epsilon\) if x is irrational. Choose \(\displaystyle \epsilon=1/2\)
If x is rational, \(\displaystyle f_n(x)-f(x)=1-1=0< \epsilon\). I think f(x)=1 because eventually \(\displaystyle i \leq n\) as n approaches infinity.
The part for x rational sounds a litte sloppy to me, apart from that I don't quite see where your problem is. To fix that sloppiness you should argue that if \(\displaystyle x\in [0;1]\cap\mathbb{Q}\), then there exists a unique \(\displaystyle n_0\) such that \(\displaystyle x=a_{n_0}\) and therefore for all \(\displaystyle n\geq n_0\) we have that \(\displaystyle |f_n(x)-1|=|f_n(a_{n_0})-1|=|1-1|=0<\epsilon\)

f is not integrable because all the lower sums are 0 and all the upper sums are 1.

2) For \(\displaystyle n \in N\) define \(\displaystyle f_n: [0,1] \rightarrow R\) by \(\displaystyle f_n(x)=1/x\) if \(\displaystyle x \in [1/n,1]\) and \(\displaystyle f_n(x)=n^2x\) if \(\displaystyle x \in [0,1/n)\). Prove that \(\displaystyle \{f_n\}\) does not converge uniformly on [0,1] but converge uniformly on [M,1] where 0<M<1.

I don't know what the pointwise limit of \(\displaystyle f_n\), so I can't go any anywhere. (Crying) It would be really helpful if someone helps me how to find the pointwise limit
What about \(\displaystyle f(x) := \begin{cases}0, &\text{if } x=0\\\tfrac{1}{x}, &\text{otherwise}\end{cases}\)
 
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