Examples in which $m(X)=∞ $ and $L^p(X,B,m)⊂L^q(X,B,m)$ for some p,q with 1⩽p<q<∞?

Question

I've found this theorem online:

Theorem: Let $(X,B,m)$ be a σ-finite measure space, where m is a non-negative measure. Then the following conditions are equivalent:

1. We have $L^p(X,B,m)⊃L^q(X,B,m)$ for some $p,q$ with $1⩽p<q<∞$.
2. $m(X)<∞$.

I read the proof, but I can't convince myself that these two conditions are equivalent. Could you , please, give me some examples in which $m(X)= ∞$ and $L^p(X,B,m)⊂ L^q(X,B,m)$ for some $p,q$ with $1⩽p<q<∞$?

Answer 1

The question is a little confusing; note that what you are asking has nothing to do with the theorem that you have seen; you should be asking for a measure space with infinite measure such that there exist $p<q\in [1,\infty)$ so that $L^q\not\subset L^p$.

Anyway, the theorem is true. I will try to demonstrate the proof in detail, using elementary functional analysis (the closed graph theorem to be more precise).

Let $1\leq p<q<\infty$ and $f\in L^q(X)$. Then we can apply Holder's inequality for the numbers $s=q/p>1$ and to its conjugate $t$ (i.e. $\frac{1}{s}+\frac{1}{t}=1)$, so $$\|f\|_p^p=\int_X|f|^pd\mu=\int_X|f|^p\cdot1d\mu\leq\bigg(\int_X(|f|^p)^{q/p}d\mu\bigg)^\frac{p}{q}\cdot\bigg(\int_X1^td\mu\bigg)^{1/t}=\|f\|_q^p\cdot\mu(X)^{1/t}<\infty $$ So $$\|f\|_p\leq\mu(X)^{1/tp}\|f\|_q=\mu(X)^{\frac{q-p}{pq}}\|f\|_q\;\;\;(\star)$$ and this is true for all functions $f\in L^q$.

$(2)\implies(1)$ If $\mu(X)<\infty$ and $1\leq p<q<\infty$ and $f\in L^q$, then $\|f\|_p\leq\mu(X)^{\frac{q-p}{pq}}\|f\|_q<\infty$ by the above inequality, so $f\in L^p$ and the claim is proved.

$(1)\implies(2)$ Suppose that $L^q\subset L^p$ for all $1\leq p<q<\infty$. We can then define the identity operator $T_{q,p}:L^q(X)\to L^p(X)$ by $f\mapsto f$. Since both $L^p$ and $L^q$ are Banach spaces, we can apply the closed graph theorem to show that $T_{q,p}$ is a bounded operator: Indeed, if $(f_n)\subset L^q$ is a sequence with $\|f_n\|_q\to0$ and $\|f_n-g\|_p\to0$ for some $g\in L^p$, then we have that $f_n$ has a subsequence $f_{n_k}$ converging to $0$ almost everywhere. But $\|f_{n_k}-g\|_p\to0$, so $f_{n_k}$ has a subsequence covnerging to $g$ almost everywhere. But we already established that $f_{n_k}$ converges to $0$ almost everywhere, so $g=0$ almost everywhere, thus $\|f_{n}\|_p\to0$, i.e. $T_{q,p}f_n\to0$ in $L^p$, and by the closed graph theorem this shows that $T_{q,p}$ is bounded. In other words, $\|T_{q,p}\|<\infty$.

Recall the definition of operator norm: if $S:X\to Y$ is a linear operator, then $$\|S\|=\inf\{c\in[0,\infty): \|Sx\|\leq c\|x\| \text{ for all }x\in X\}=\sup_{x\in X,x\neq0}\frac{\|Sx\|}{\|x\|}\;\;(\star\star)$$ By $(\star)$ and the first equation of $(\star\star)$, $\|T_{q,p}\|\leq\mu(X)^{\frac{q-p}{pq}}$. On the other hand, if $f$ is the constant function $1$ on all $X$, then by the second equation of $(\star\star)$ we have that $$\|T_{q,p}\|\geq\frac{\|f\|_p}{\|f\|_q}=\mu(X)^{\frac{1}{p}-\frac{1}{q}}=\mu(X)^{\frac{q-p}{pq}} $$

Combining with the above, $$\|T_{q,p}\|=\mu(X)^{\frac{q-p}{pq}}$$ and since $\|T_{q,p}\|<\infty$ we have that $\mu(X)<\infty$.