$L^q[0,1]$ is a (Baire) first category subset in $L^p[0,1]$

Question

Let $1\leq p<q<\infty$, prove that $L^q[0,1]$ is a (Baire) first category subset in $L^p[0,1]$

There is an answer to this question here. In that answer, I figured out that there are two steps:

(1) $\overline{B}_q(0,c)$ is closed in $L^p[0,1]$;

(2) $\overline{B}_q(0,c)$ has empty interior in $L^p[0,1]$.

But I need to prove it by applying Baire category theorem and uniform boundedness principle respectively. (Because I just learned Baire category theorem and uniform boundedness principle in the functional analysis lesson. )

I have no idea how to construct a family of bounded linear operator and what to prove by using uniform boundedness principle. And I don't know how to use Baire category theorem either.

Could you please give some hints?

Answer 1

We will use the following formulation of the uniform boundedness theorem.

Theorem 1. Let $X$ and $Y$ be normed spaces and $\mathcal{F} \subseteq \mathcal{L}(X,Y)$ a non-empty family of bounded linear operators from $X$ into $Y$. Suppose the set \begin{equation} \{x\in X : \{T(x) : T \in \mathcal{F} \} \text{ is bounded} \} \end{equation} contains a set of the second category in $X$. Then $\{T(x) : T \in \mathcal{F} \}$ is bounded for all $x\in X$. Moreover, the family $\mathcal{F}$ is uniformly bounded, that is, there exists some $M \geq 0$ such that $\|T\|_{\mathcal{L}(X,Y)} \leq M$ for all $T \in \mathcal{F}$.

The standard proof of Theorem 1 makes use of the Baire category theorem. You can see 5.13 of Real Analysis by Folland for a proof of the uniform boundedness theorem in the above formulation.

We will also use the following elementary result. A proof can be found in Theorem 6.14 of Real Analysis by Folland.

Lemma 1. Let $(X, \mathcal{A}, \mu )$ be a measure space. Let $p\in [1, \infty ]$ and let $p'$ denote the Hölder conjugate exponent of $p$. Let $g \colon X \to \mathbb{C}$ be measurable. Suppose the following conditions hold.

1. The set $\{x\in X : g(x) \neq 0 \}$ is $\sigma$-finite.
2. We have $fg \in L^{1}(X, \mu )$ for all $f\in \Sigma$, where $\Sigma$ denotes the collection of simple functions $u\colon X \to \mathbb{C}$ such that the set $\{x\in X : u(x) \neq 0\}$ has finite measure.
3. The set \begin{equation} \Big\{ \Big| \int_{X} fg \, d\mu \, \Big| : f\in \Sigma \text{ and } \|f\|_{L^{p'}} = 1 \Big\} \tag{$*$} \end{equation} is bounded.

Then $g\in L^{p}(X, \mu )$ and the supremum of ($*$) is equal to $\|g\|_{L^{p}}$.

We now address the question.

Theorem. Let $p,q\in [1, \infty ]$ with $p < q$. Then $L^{q}[0,1]$ is of the first category in $(L^{p}[0,1], \|\cdot\|_{L^{p}})$.

Proof. Let $A$ denote the collection of simple functions $u\colon [0,1] \to \mathbb{C}$ such that $\|u\|_{L^{q'}} = 1$. For each $f\in A$ define $\phi_{f} \colon L^{p}[0,1] \to \mathbb{C}$ by $\phi_{f}(g) := \int_{[0,1]} fg$. Since $A \subseteq L^{p'}[0,1]$, it follows from the Hölder inequality that $A \subseteq \mathcal{L}(L^{p}[0,1], \mathbb{C})$.

Let $g\in L^{p}[0,1]$. We have \begin{equation} \{ \phi_{f}(g) : f\in A \} = \Big\{ \int_{[0,1]} fg : f\in A \Big\} . \tag{1} \end{equation} It follows from $(1)$, Lemma 1 and the Hölder inequality that \begin{equation} \{g\in L^{p}[0,1] : \{ \phi_{f}(g) : f\in A \} \text{ is bounded} \} = L^{q}[0,1]. \tag{2} \end{equation} Since $L^{q}[0,1] \neq L^{p}[0,1]$, we deduce from $(2)$ and Theorem 1 that $L^{q}[0,1]$ is of the first category in $L^{p}[0,1]$.