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Let $f_k: \mathbb{R} \rightarrow \mathbb{R} , f_k(x) = \frac{1}{\sqrt{x}}\chi_{\left[\frac{1}{2^{k+1}},\frac{1}{2^k}\right]}(x)$

For $k \rightarrow \infty $ the interval $[\frac{1}{2^{k+1}},\frac{1}{2^k}]$ will decrease it's range, but I would say that for every $k \in \mathbb{N}$ there exists an $x \in \mathbb{R}$ so that $x \in [\frac{1}{2^{k+1}},\frac{1}{2^k}]$.

Can anyone tell me the function $f \leftarrow f_k$ to which the sequence converges and how? Pointwise of course, even of course not. But in measure or in $L^1, L^2 ,...$?

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You need to fix $x>0$. Then there is $K$ such that $\frac{1}{2^k}<x$, for $k>K$. Then $f_k(x)=0$, for $k>K$. Then $f(x)=0$.

For $x\leq0$, $f_k(x)=0$, for all $k$. Then $f(x)=0$.

Now, for $L^1$ we can integrate.

$$||f_k||_1=\int_{1/2^{k+1}}^{1/2^{k}}x^{-1/2}=\left.\frac{x^{1-1/2}}{1-1/2}\right|_{1/2^{k+1}}^{1/2^k}=2\left(\frac{1}{2^{k/2}}-\frac{1}{2^{(k+1)/2}}\right)\to0$$

This tells us that $||f_k-0||_1\to0$.

We can do the computation in general, in $L^p$:

$$||f_k||_p^p=\int_{1/2^{k+1}}^{1/2^{k}}x^{-p/2}=\begin{cases}\left.\frac{x^{1-p/2}}{1-p/2}\right|_{1/2^{k+1}}^{1/2^k}=\frac{1}{1-p/2}\left(\frac{1}{2^{k(1-p/2)}}-\frac{1}{2^{(k+1)(1-p/2)}}\right)&\text{ for }p\neq2\\\log(x)|_{1/2^{k+1}}^{1/2^k}=\log(1/2^k)-\log(1/2^{k+1})=\log(2)&\text{ for }p=2\end{cases}$$

The difference is now, that if $1-p/2\leq0$ then the limit doesn't go to zero.

We know that for a convergent sequence in $L^p$ we can always extract a subsequence that converges pointwise a.e. Therefore if it were to converge it would converge to $f=0$. Therefore, for $1-p/2\leq0$ it doesn't converge. For $1-p/2>0$ it tends to $f=0$ as before.

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  • $\begingroup$ Wow, that was a quick sufficiant answer, thank you very much! And I guess I can assume that $f_k$ converges in measure, because $||f_k - f|| = ||f_k|| \rightarrow 0$, right? Or does that follow directly out of $L^1$ convergence? $\endgroup$ – RoyPJ Jan 28 '15 at 14:49
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    $\begingroup$ The set of points where $|f_k(x)-f(x)|>\epsilon$ is the interval $[1/2^{k+1},1/2^k]$. Its size is $\frac{1}{2^k}-\frac{1}{2^{k+1}}\to0$. $\endgroup$ – Pp.. Jan 28 '15 at 14:51
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Your sequence is going to zero pointwise. On the one hand, $f_k(0) \equiv 0$. On the other hand, if $x>0$ and $k>-\log_2(x)$, then $f_k(x)=0$.

Additionally, your sequence is dominated by $f(x) = \frac{1}{\sqrt{x}} \chi_{[0,1]}(x)$, which is in $L^p$ for $p \in [1,2)$. So it is definitely going to zero in $L^p$ for $p \in [1,2)$. It does not go to zero in $L^2$: by direct calculation we have $\| f_k \|_2^2=\int_{2^{-k-1}}^{2^-k} \frac{1}{x} dx=\ln(2^{-k})-\ln(2^{-k-1})=\ln(2)$. This means it does not go to zero in any $L^p$ for $p>2$ either (because on a finite measure space, convergence in $L^{p_1}$ implies convergence in $L^{p_2}$ if $p_2 \leq p_1$).

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