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?

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That would depend on what $A_n,B_n$ would be. –  tomasz Oct 29 '12 at 2:03
@tomasz I don't follow... if need be, ignore $A_n$ and $B_n$ and just consider $\Vert A-B\Vert_3$ (and similarly, 2 and 1) –  fkmasö Oct 29 '12 at 2:21
Presumably, you are getting at this: math.stackexchange.com/questions/21460/… –  JavaMan Oct 29 '12 at 17:35

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

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

2. ${L^{q}}(X,\Sigma,\mu) \supseteq {L^{p}}(X,\Sigma,\mu)$ 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.

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