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Consider the interval $I = [0,1]$ and the sequence of functions:

$$f_n(x) = (-1)^k \ \text{for} \displaystyle \frac{k}{2^n} \le x < \frac{k+1}{2^n} \ \text{where} \ 0 \le k < 2^n - 1$$

I want to exhibit that $f_n \not \to f \equiv 0$ strongly in $L^p[I]$ $\forall 1 \le p < \infty$

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What is $\|f_n\|_p$? – Jonas Meyer Feb 14 '13 at 4:33
Have you tried sketching the first few $f_n$’s? – Brian M. Scott Feb 14 '13 at 4:35
up vote 1 down vote accepted

The definition of the $L_p$ norm of $f_n$ is $\|f_n\|_p=\left(\int_I |f_n(x)|^pdx\right)^{1/p}$. If $y=1$ or $y=-1$, then $|y|^p=1$. What is $\|f_n\|_p$?

The definition of convergence of a sequence $(f_n)$ to zero in $L^p(I)$ says that $\lim\limits_{n\to\infty}\|f_n\|_p=0$. Does this happen for $(f_n)$ in your question?

The sequence does converge weakly to $0$.

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It seems to me that $\| f_n\|_p = 1$ always hence the limit cannot be 0. Also, I amended the question. I was talking about strong convergence. Sorry for the confusion! – user44069 Feb 14 '13 at 4:57
@Stefan: Yes, that is what I thought. I was not confused, I just wanted to add the remark about the case of weak convergence because I think it is interesting and related; I suspect it is the motivation for cooking up this example. I am glad you found $\|f_n\|_p$. – Jonas Meyer Feb 14 '13 at 5:06

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