Prove or disprove: If there exists a positive constant $M$ such that $||f||_{L^p(\mathbb{R})}<M$ for any $p > 1$, then $f \in L^1(\mathbb{R})$.
We have $$\left(\int_{\mathbb{R}}|f|^p\right)^{1/p} < M \implies \int_\mathbb{R}|f|^p < M^p.$$ Thus $f \in L^p(\mathbb{R})$. But since $1 < p$, wouldn't it automatically mean that $f \in L^1(\mathbb{R})$?
Anyway, I'm thinking of using Holder's inequality to prove that the statement is true, since on the left side we will always have $||f||_1$. Let $1<p,p'<\infty$ such that $\frac{1}{p}+\frac{1}{p'}=1$, then we have $$||f||_1 \leq ||f||_p||f||_{p'}$$ and so $$\int _\mathbb{R}|f| \leq \left(\int_\mathbb{R}|f|^p\right)^{\frac{1}{p}}\left(\int_\mathbb{R}|f|^{p'}\right)^{\frac{1}{p'}}<M\cdot M = M^2.$$ Hence $f \in L^1(\mathbb{R})$. Did I miss out anything?
EDIT. I forgot that the left side should be a product of functions, let me rewrite it as: $$\int_\mathbb{R}f^2 < M^2$$ which doesn't seem helpful. Also, $\mathbb{R}$ has infinite measure. I'm starting to think that this statement is false.
This is an immediate consequence of Fatou's Lemma. We have $\int |f|^{p} \leq M^{p}$. Letting $p \to 1$ ,say through the sequence $(1+\frac 1 n)$, we get $\int |f| \leq \lim \inf \int |f|^{p}\leq \lim \inf M^{p}=M$.