So this is a question on a final exam I just had.
Let $X\subset L^1_m\left([0,1]\right)$ be a closed subspace of the Banach space $L^1_m\left([0,1]\right)$. Assume that $$\forall f\in X,\exists p\ \text{s.t.}\ f\in L^p_m\left([0,1]\right)$$ then show that $\exists p_0>1$ s.t. $X\subset L^{p_0}_m\left([0,1]\right)$.
I already know of a solution using Baire Category Theorem, but I went down a route using more "elementary" constructions, and I want to know if there's a way I can finish off my proof. This is what I have so far:
Assume not, so there is a strictly decreasing sequence $\{p_n\}$ of real numbers converging to $1$ so that there is a function $f_n\in X$ where $f_n\in L^{p_n}$ but $f \not \in L^{p_{n+1}}$. Then define $$g_n:=\sum_{i=1}^n\frac{f_i}{2^i\cdot ||f_i||_1}$$. Because $X$ is a subspace, we know that $g_n\in X$ $\forall n$. Furthermore, we have that $$||g_{n+k}-g_n||_1=||\sum_{i=n}^{n+k}\frac{f_i}{2^i\cdot ||f_i||_1}||_1\leq \sum_{i=n}^{n+k}\frac{||f_i||_1}{2^i\cdot ||f_i||_1}=\sum_{i=n}^{n+k}\frac{1}{2^i}$$ so taking $n\to \infty$ we can minimize this arbitrarily. Thus we have a Cauchy sequence, and since a closed subspace of a Banach space is a Banach space, we know that there is some $g\in X$ so that $g_n\to g$ in $L^1$. What I want to argue at this point is that $g\not \in L^p$ for any $p>1$. I've tried for hours playing around with different subsequences and inequalities but I'm coming up short. Please let me know if I should abandon this path, or if there is a way to make this work.