# Find a function $f$ such that $f \in L_p$ if and only if $1 ≤ p ≤ p_0$ with certain measure.

I am working on this question which is from Bartle's "The Elements of Integration and Lebesgue Measure" (questions 6.H and 6.I for reference).

Let $$X = \mathbb{N}$$ and let $$\mathbb{X}$$ be the collection of all subsets of N. Let $$λ$$ be defined by:

\begin{align*} λ(E) = 􏰊\sum_{n=1}^{\infty} \frac{1}{n^2}, E \in \mathbb{X}, n \in E \end{align*}

$$(a)$$ Show that $$f : X → R, f(n) = \sqrt{n}$$ satisfies $$f \in L_p$$ if and only if $$1 ≤ p < 2$$.

$$(b)$$ Find a function $$f$$ such that $$f \in L_p$$ if and only if $$1 ≤ p ≤ p_0$$.

Part $$(a)$$ is simple enough by the divergence of harmonic series as;

\begin{align*} \int |f|^p d\lambda = 􏰊\sum_{n=1}^{\infty} \frac{|f(n)|^p}{n^2} \end{align*}

From that it is also easy to show that, again by divergence of harmonic series, that for $$f(n)=n^\frac{1}{p_o}$$ if and only if $$1 ≤ p < p_0$$.

I am stuck on trying to extend this to include when $$p=p_0$$. My first thought is to somehow change the function $$f(n)=n^\frac{1}{p_o}$$ such that series is the alternating harmonic series when $$p = p_0$$ and this still convergent, however this is tricky due to power involved. Any tips would be much appreciated.

Here's another idea: Let $$(p_k)_{n=1}^{\infty}$$ be a sequence in $$(p_0,\infty)$$ such that $$p_k \to p_0$$. Define functions $$f_k(n) = n^{1/p_k}$$ and let $$f(n) = \sum_{k=1}^{\infty}\frac{1}{2^k\Vert f_k \Vert_{p_0}^{p_0}}f_k(n).$$ By what you have shown, $$f_k \in L^p$$ if and only if $$1 \leq p < p_k$$. In particular $$f_k \in L^{p_0}$$. Since $$f_k(n) \geq 1$$, for $$1 \leq p \leq p_0$$ we have $$\Vert f_k \Vert_p^p \leq \Vert f_k \Vert_{p_0}^{p_0}.$$ Taking $$p$$-th roots and noting that $$\Vert f_k \Vert_{p_0} > 1,$$ we see that $$\Vert f_k \Vert_p \leq \Vert f_k \Vert_{p_0}^{p_0/p} \leq \Vert f_k \Vert_{p_0}^{p_0},$$ thus $$\Vert f \Vert_p \leq \sum_{k=1}^{\infty} \frac{1}{2^k \Vert f_k \Vert_{p_0}^{p_0}} \Vert f_k \Vert_p \leq \sum_{k=1}^{\infty} \frac{1}{2^k} < \infty,$$ so $$f \in L^p$$ for $$1 \leq p \leq p_0$$. On the other hand, if $$p > p_0$$, then choosing $$k$$ such that $$p_k < p$$ (here we use $$p_k \to p_0$$), we have that $$f_k \not\in L^p$$. Then $$f \not\in L^p$$ since $$f \geq \frac{1}{2^k \Vert f_k \Vert_{p_0}^{p_0}} f_k.$$