Why is $L^3$ weaker than $L^2$? Someone told me today that if I can show $\Vert A_n-B_n\Vert_3\to 0$ as $n\to \infty$, then claiming $A=B$ as $n\to \infty$ (where $A$ and $B$ are the respective limits of $A_n$ and $B_n$) is a weaker claim than if I were to show that $\Vert A_n-B_n\Vert_2\to 0$ (which in turn is weaker than if I were to show $\Vert A_n-B_n\Vert_1\to 0$). Why is this so?
 A: The answer pretty much depends on the measure space $ (X,\Sigma,\mu) $ that you are working with. There is the following well-known result (Rudin poses it as an exercise in his Real and Complex Analysis):
Let $ 1 \leq p < q \leq \infty $. Then


*

*$ {L^{q}}(X,\Sigma,\mu) \subseteq {L^{p}}(X,\Sigma,\mu) $ iff $ X $ does not have measurable subsets of arbitrarily large (but non-infinite) measure.

*$ {L^{q}}(X,\Sigma,\mu) \supseteq {L^{p}}(X,\Sigma,\mu) $ iff $ X $ does not have measurable subsets of arbitrarily small (but non-zero) measure.
If $ \mu(X) < \infty $, in which case (1) holds, we even have the following relationship between the norms $ \| \cdot \|_{p} $ and $ \| \cdot \|_{q} $:
$$
\forall \, \text{measurable functions $ f $ on $ X $}: \quad \| f \|_{p} \leq [\mu(X)]^{\frac{1}{p} - \frac{1}{q}} \cdot \| f \|_{q}.
$$
This yields the following sequence of implications:
$$
\lim_{n \rightarrow \infty} \| f_{n} \|_{3} = 0 \quad \Longrightarrow \quad \lim_{n \rightarrow \infty} \| f_{n} \|_{2} = 0 \quad \Longrightarrow \quad \lim_{n \rightarrow \infty} \| f_{n} \|_{1} = 0.
$$
I do not know what the conditions on $ (X,\Sigma,\mu) $ have to be in order to obtain the reverse of the above sequence of implications, namely,
$$
\lim_{n \rightarrow \infty} \| f_{n} \|_{1} = 0 \quad \Longrightarrow \quad \lim_{n \rightarrow \infty} \| f_{n} \|_{2} = 0 \quad \Longrightarrow \quad \lim_{n \rightarrow \infty} \| f_{n} \|_{3} = 0.
$$
This is the case that you are interested in. There may be results in interpolation theory (similar to the Riesz-Thorin Interpolation Theorem) that would give you precisely what you need.
