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Hmmm...I am almost embarrassed to ask this question, but I'll ask anyway. How do I show that the sequence defined by $f_n(x) = n^{1/p}\chi_{[0,1/n]}$ ,$1\le p \lt \infty$ and $x\in [0,1]$ converges pointwise to $0$.

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Are you sure that you have it right? Are you sure it's $n^{1/p}$ and not $x^{1/p}$? – Asaf Karagila Apr 20 '12 at 18:39
Im pretty sure it's $n^{1/p}$. – Linda Apr 20 '12 at 18:52
I am guessing that the purpose for studying this sequence of functions is that they all have $L^p$-norms equal to one. IOW, I believe in that $n^{1/p}$ and, consequently, we have pointwise convergence only for $x\in(0,1]$. – Jyrki Lahtonen Apr 20 '12 at 19:36
up vote 3 down vote accepted

It still doesn't, I'm afraid. We have that $f_n(0)=n^{1/p}\chi_{[0,\frac{1}{n}]}(0)=n^{1/p}$ for all $n$, and hence $$\lim_{n\to\infty}f_n(0)=\lim_{n\to\infty} n^{1/p}= \infty.$$ However, for any $x<0$, we have that $f_n(x)=\chi_{[0,\frac{1}{n}]}(x)=0$ for all $n$, and for any $x>0$ there is some $m$ for which $x>\frac{1}{m}$, hence $x\notin [0,\frac{1}{n}]$ for all $n>m$, hence $\chi_{[0,\frac{1}{n}]}(x)=0$ for all $n>m$.

Thus the pointwise limit is the function to the extended reals $$f(x)=\begin{cases}\infty &\text{ if }x=0,\\0 &\text{ if }x\neq0. \end{cases}$$

If it's as Asaf is proposing, and the original question was actually about $f_n(x)=x^{1/p}\chi_{[0,\frac{1}{n}]}$, then just note that $$f_n(0)=0\cdot \chi_{[0,\frac{1}{n}]}(0)=0\cdot 1=0$$ for all $n$, so that you do have $$\lim_{n\to\infty}f_n(0)=\lim_{n\to\infty} 0= 0.$$ Combined with my earlier argument for $x\neq0$, this then shows that the pointwise limit of $f_n(x)=x^{1/p}\chi_{[0,\frac{1}{n}]}$ is the zero function.

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I am so sorry. I forgot the $n^{1/p}$ part – Linda Apr 20 '12 at 18:28
I think it still doesn't converge pointwise to 0. – Zev Chonoles Apr 20 '12 at 18:36
Does it converge to $0$ almost everywhere? – Linda Apr 20 '12 at 18:53
Yes, since $\{0\}$ is a set of measure 0, the function $$f(x)=\begin{cases}\infty &\text{ if }x=0,\\0 &\text{ if }x\neq0. \end{cases}$$ differs from the zero function on a set of measure zero, hence the pointwise limit function is almost everywhere 0. – Zev Chonoles Apr 20 '12 at 18:54

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