Let $\phi:\Bbb R\to\Bbb R$ be a function given by: $$\phi(x)= \left\{ \begin{array}{ll} x^{-1/2} & x\in (0,1) \\ 0 & x\notin (0,1) \\ \end{array} \right.$$

Let $\sum_{k=1}^\infty a_k$ be a convergent series of real positive numbers. Let $\Bbb Q=\{r_1,r_2,...\}$ be an enumeration. Let $f:\Bbb R\to [0,\infty)$ be given by $f(x)=\sum_{k=1}^\infty a_k \phi(x-r_k)$, then the next two results follows:

$$i)\;\ f\in L_1(\lambda)\\ ii)\;\ f\notin L_2(a,b)\;\ \;(\text{Hint: density of $\Bbb Q$ in $\Bbb R$})$$

Proof: (what I've done)

For the $i)$ part:

By definition, its clear that $\forall x\in \Bbb R:\ \ \phi(x)\ge 0\;\ $and $\;\ \phi(x)=x^{-1/2} \mathbb{1}_{(0,1)}(x)\ $. So let $c\in \Bbb R$: $$\Rightarrow \phi^{-1}([c,\infty))= \left\{ \begin{array}{ll} \Bbb R & c\le 0 \\ (0,1) & c\in (0,1) \\ (0,\frac{1}{c^2}] & c\ge 1 \end{array} \right.\\ \Rightarrow\;\ \phi^{-1}([c,\infty))\in \Bbb B_{\Bbb R}\;\ \Rightarrow\;\ \phi\in M(\Bbb R,\Bbb B_{\Bbb R})$$

Now, let $S_n=\sum_{k=1}^n a_k$. Clearly $S_n$ is measurable $\forall n\in \Bbb N$, since $S_n$ is a finite linear combination of constants ($a_k$) for every $n\in \Bbb N$

Then let $\varphi(x)=\phi(x-r_k)\;\ \forall x\in \Bbb R\;\ \forall k\in \Bbb N$. And define: $$\;\ f_n=S_n \varphi\;\ \forall n\in \Bbb N\Rightarrow f_n\in M(\Bbb R,\Bbb B_{\Bbb R})\;\ \forall n\in \Bbb N$$

And, since $\varphi\ge 0$ and $S_n\le S_{n+1}\;\ \forall n\in N$, $(f_n)_{n\in \Bbb N}\;\ $is a non-decreasing sequence such that:

$$\lim_{n\to \infty} f_n(x)=\lim_{n\to \infty}S_n \varphi(x)=\lim_{n\to \infty}S_n \phi(x-r_k)=\sum_{k=1}^\infty a_k \phi(x-r_k)=f(x)\;\ \forall x\in \Bbb R$$

So, applying the MCT we get that: $$\lim_{n\to \infty} \int_{B}S_n \varphi d\lambda =\int_{B}fd\lambda\quad \forall B\subseteq \Bbb R$$

So for $B=\Bbb R$ we get that:

$$\int fd\lambda=\lim_{n\to \infty} \int S_n \varphi d\lambda=\lim_{n\to \infty}S_n \int \varphi d\lambda\\ \text{where}\;\ \varphi(x)=\phi(x-r_k)=(x-r_k)^{-1/2} \mathbb{1}_{(r_k,r_k+1)}(x)\;\ \forall x\in \Bbb R\\ \Rightarrow\;\ \int fd\lambda=\lim_{n\to \infty}S_n\int_{(r_k,r_k+1)}(x-r_k)^{-1/2} d\lambda(x)=\lim_{n\to \infty}S_n\int_{r_k}^{r_k+1}(x-r_k)^{-1/2} d(x)\\ =\lim_{n\to \infty}S_n\ 2 \Big[ (r_{k}+1-r_k)^{1/2} - (r_{k}-r_k)^{1/2} \Big]=2\lim_{n\to \infty}S_n=2\sum_{k=1}^\infty a_k<\infty $$

Thus $f\in L_1(\lambda)$ (I think its clear, but would appreciate if you think something is missing)

For the $ii)$ part:

I need to prove that $\int_{(a,b)} |f|^2 d\lambda = \infty$. So I started with $|f|^2$:

$$|f(x)|^2 = \Big[ \lim_{n\to \infty} \sum_{k=1}^n a_k\phi(x-r_k) \Big]^2=\lim_{n\to \infty} \Big[ \sum_{k=1}^n a_k\phi(x-r_k) \Big]^2\ge \lim_{n\to \infty} \Big(\sum_{k=1}^n a_k\Big)^2 \phi^2(x-r_k)\\ \ge \lim_{n\to \infty} \sum_{k=1}^n a_k^2\ \phi^2(x-r_k)\ge \lim_{n\to \infty} \sum_{k=1}^n a_k\ \phi^2(x-r_k)\;\ \forall x\in \Bbb R\\ \text{where all the inequalities follow because we have finite sums of positive elements}$$

So $|f(x)|^2 \ge \lim_{n\to \infty} \sum_{k=1}^n a_k\ \phi^2(x-r_k)\;\ \forall x\in \Bbb R,\;\ $where $\Big(\sum_{k=1}^n a_k\ \phi^2(x-r_k)\Big)_{n\in \Bbb N}$ is clearly a measurable and non-decreasing sequence such that converges to $\sum_{k=1}^\infty a_k\ \phi^2(x-r_k)$, thus applying MCT for $B=(a,b)\subset \Bbb R$:

$$\int_{(a,b)} \sum_{k=1}^\infty a_k\ \phi^2(x-r_k)\ d\lambda(x)=\lim_{n\to \infty} \int_{(a,b)} \sum_{k=1}^n a_k\ \phi^2(x-r_k)\ d\lambda(x)\\ =\lim_{n\to \infty} \sum_{k=1}^n a_k\ \int_{(a,b)} \phi^2(x-r_k)\ d\lambda(x)=\lim_{n\to \infty} \sum_{k=1}^n a_k\ \int_{(a,b)} \frac{1}{(x-r_k)} \mathbb{1}_{(r_k,r_k+1)}(x)\ d\lambda(x)$$

And got stuck here since I don't get how the density of $\Bbb Q$ in $\Bbb R$can help. I don't know how's the process to justify that:

$$\lim_{n\to \infty} \sum_{k=1}^n a_k\ \int_{(a,b)} \frac{1}{(x-r_k)} \mathbb{1}_{(r_k,r_k+1)}(x)\ d\lambda(x)=\lim_{n\to \infty} \sum_{k=1}^n a_k\ \int_{r_k}^{r_k+1} \frac{1}{(x-r_k)} d(x)=\lim_{n\to \infty} \sum_{k=1}^n a_k\ \Big[ log(r_k+1-r_k) - \lim_{x\to r_k} log(x-r_k) \Big]\\ \text{where}\;\ - \lim_{x\to r_k} log(x-r_k)\to \infty\\ \text{and, thus}\;\ \int_{(a,b)}|f|^2\ d\lambda=\infty$$ Any help will be appreciated.


I think I got it. So, we've already proved that:

$$|f(x)|^2\ge \lim_{n\to \infty} \sum_{k=1}^n a_k \phi^2(x-r_k)\;\ \forall x\in \Bbb R$$


$$\lim_{n\to \infty} \sum_{k=1}^n a_k \phi^2(x-r_k)=\sum_{k=1}^\infty a_k \phi^2(x-r_k)\ge \sum_{k=1}^n a_k \phi^2(x-r_k)\;\ \forall n\in \Bbb N\;\ \forall x\in \Bbb R$$


$$\int_{(a,b)} |f|^2d\lambda\ge \int_{(a,b)} \sum_{k=1}^n a_k \phi^2(x-r_k)\ d\lambda(x)=\sum_{k=1}^n a_k\int_{(a,b)} \phi^2(x-r_k)\ d\lambda(x)\\ =\sum_{k=1}^n a_k\int_{(a,b)} \frac{1}{x-r_k}\mathbb{1}_{(r_k,r_k+1)}(x)\ d\lambda(x)=\sum_{k=1}^n a_k\int_a^b \frac{1}{x-r_k}\mathbb{1}_{(r_k,r_k+1)}(x)\ dx\\ \Rightarrow \int_{(a,b)} |f|^2d\lambda\ge \sum_{k=1}^n a_k\int_a^b \frac{1}{x-r_k}\mathbb{1}_{(r_k,r_k+1)}(x)\ dx$$

Now, since $\Bbb Q$ is dense in $\Bbb R$ then: $$\exists\ r_{k_0}\in \Bbb Q\;\ \text{s.t.}\;\ a<r_{k_0}<b$$

and $$\sum_{k=1}^n a_k\int_a^b \frac{1}{x-r_k}\mathbb{1}_{(r_k,r_k+1)}(x)\ dx\ge a_{k_0}\int_a^b \frac{1}{x-r_{k_0}}\mathbb{1}_{(r_{k_0},r_{k_0}+1)}(x)\ dx$$

so $$\int_{(a,b)} |f|^2d\lambda\ge a_{k_0}\int_a^b \frac{1}{x-r_{k_0}}\mathbb{1}_{(r_{k_0},r_{k_0}+1)}(x)\ dx$$

where $$a_{k_0}\int_a^b \frac{1}{x-r_{k_0}}\mathbb{1}_{(r_{k_0},r_{k_0}+1)}(x)\ dx=a_{k_0}\Big[\int_a^{r_{k_0}} \frac{1}{x-r_{k_0}}\mathbb{1}_{(r_{k_0},r_{k_0}+1)}(x)\ dx + \int_{r_{k_0}}^{b} \frac{1}{x-r_{k_0}}\mathbb{1}_{(r_{k_0},r_{k_0}+1)}(x)\ dx\Big]\\ =a_{k_0} \int_{r_{k_0}}^{b} \frac{1}{x-r_{k_0}}\mathbb{1}_{(r_{k_0},r_{k_0}+1)}(x)\ dx$$

So, if $b<r_{k_0}+1$: $$a_{k_0} \int_{r_{k_0}}^{b} \frac{1}{x-r_{k_0}}\mathbb{1}_{(r_{k_0},r_{k_0}+1)}(x)\ dx=a_{k_0} \int_{r_{k_0}}^{b} \frac{1}{x-r_{k_0}}\ dx=a_{k_0}\Big[log(b-r_{k_0})-\lim_{x\to r_{k_0}} log(x-r_{k_0})\Big]\to \infty$$

and, if $b>r_{k_0}+1$: $$a_{k_0} \int_{r_{k_0}}^{b} \frac{1}{x-r_{k_0}}\mathbb{1}_{(r_{k_0},r_{k_0}+1)}(x)\ dx=a_{k_0}\Big[ \int_{r_{k_0}}^{r_{k_0}+1} \frac{1}{x-r_{k_0}}\ dx+\int_{r_{k_0}+1}^{b} \frac{1}{x-r_{k_0}}\cdot 0\ dx \Big]\\ =a_{k_0}\int_{r_{k_0}}^{r_{k_0}+1} \frac{1}{x-r_{k_0}}\ dx=a_{k_0}\Big[log(1)-\lim_{x\to r_{k_0}} log(x-r_{k_0})\Big]\to \infty$$

Thus, $$\int_{(a,b)} |f|^2d\lambda\ge a_{k_0}\int_a^b \frac{1}{x-r_{k_0}}\mathbb{1}_{(r_{k_0},r_{k_0}+1)}(x)\ dx\to \infty\\ \text{i.e.}\;\ f\notin L_2(a,b)$$

What do you think?

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