Does the function $f(x)=\frac{1}{\sqrt x}$ belong to $L^p( \mathbb N , P(\mathbb N), \mu),p=1,2,\infty?$

Question

Does the function $f(x)=\frac{1}{\sqrt x}$ belong to $L^p( \mathbb N , P(\mathbb N), \mu),p=1,2,\infty?$

$\mathbb N$- set of natural numbers, $P(\mathbb N)$- the partitive set of natural numbers. I did the question, but am very unsure if I did this correctly for $L^1,\text{ and } L^2..$

I tried this:

First off: $\sup_{esse}f(x)=M: \mu(\{x:\forall \varepsilon,|f(x)|> M + \varepsilon\})=0$

Questioning $f \in L^{\infty}( \mathbb N , P(\mathbb N), \mu): \sup_{esse}|f(x)|=\sup_{esse:n\in \mathbb N} \frac{1}{\sqrt{x}}=1 \implies f \in L^{\infty}( \mathbb N , P(\mathbb N), \mu)$ $f \in L^{1}( \mathbb N , P(\mathbb N), \mu): \sum_{1}^{\infty}\frac{1}{\sqrt{x}}=\infty \implies f \notin L^{1}( \mathbb N , P(\mathbb N), \mu)$ $f \in L^{2}( \mathbb N , P(\mathbb N), \mu): (\sum_{1}^{\infty}(\frac{1}{\sqrt{x}})^2)^{\frac{1}{2}}=\infty \implies f \notin L^{2}( \mathbb N , P(\mathbb N), \mu)$

But $$L^{\infty} \subseteq L^{2} \subseteq L^{1}$$ this we prove in class. Something doesn't add up here. Thoughts?

Answer 1

If $(\Omega,\mathcal{T},\mu)$ is a measured space such that $\mu(\Omega)<\infty$ then you have $$\forall (p,q)\in[1,\infty], p\leq q\Rightarrow L^q\subset L^p \textrm{ (with continuous injections)}$$ This doesn't apply here as $\mu(\mathbb{N})=\infty$ if $\mu$ is the usual counting measure.

However, denoting $\ell^p=L^p(\mathbb{N},\mathcal{P}(\mathbb{N}),\mu)$, you can check that $$\forall (p,q)\in[1,\infty], p\leq q\Rightarrow \ell^p\subset \ell^q \textrm{ (with continuous injections)},$$ which is the opposite inclusion! More precisely, you can prove $$\forall (p,q)\in[1,\infty], p\leq q\Rightarrow \|\cdot\|_q\leq\|\cdot\|_p.$$

Answer 2

Your conclusions about the function $f$ are correct. The inclusions are generally false (they are true for finite measure spaces).