Behavior of $\|f\|_p$ when $p$ is near the endpoint of $I$

functional-analysislp-spacesnormed-spacesreal-analysis

Given measure space $(X, \mu)$ and measurable function $f$. If $p \in (0, +\infty]$, below are the facts I know about $\Vert f \Vert_p$:

  1. (log convex in $\frac 1 p$) $\forall p_1, p_2\in (0, +\infty] \text{ such that } p_1 < p_2. \forall t \in(0, 1), \text{ let } \frac 1 p = \frac t {p_1} + \frac {1-t} {p_2}$, then $\Vert f \Vert_p \le \Vert f \Vert_{p_1}^t\Vert f \Vert_{p_2}^{1-t}$ ($0\cdot(+\infty)$ is defined to be $0$).
  2. (lower semi-continuity) $\forall p_0 \in (0, +\infty], \liminf_{p\to p_0}\Vert f\Vert_p \ge \Vert f \Vert_{p_0}$.
  3. If $0<\mu(X) < +\infty$, then $ \Vert f \Vert_p/\mu(X)^{\frac 1 p}$ is monotone increasing in $p$, combine it with (2) we get $\lim_{p\to p_0-}\Vert f\Vert_p = \Vert f \Vert_{p_0}$.

From (1) we know the set of $p$ such that $\Vert f \Vert_p < +\infty$ is an interval $I$ (possibly empty or a singleton).

  1. If $I \not\in a = \inf I$ and $a \neq 0$, from (2) we know $\lim_{p\to a+} \Vert f \Vert_p = +\infty$. If $I \not\in b = \sup I$, then $\lim_{p\to b-} \Vert f \Vert_p = +\infty$.

  2. From (1) and the dominated convergence theorem we know that if $I \ni p_0 \neq \inf I$ and $p_0 < +\infty$, then $\lim_{p\to p_0+} \Vert f \Vert_p = \Vert f \Vert_{p_0}$. If $I \ni p_0 \neq \sup I$, then $\lim_{p\to p_0-} \Vert f \Vert_p = \Vert f \Vert_{p_0}$.

  3. If $I$ is nondegenerate and $+\infty \in I$, then $\lim_{p\to +\infty}\|f\|_p = \|f\|_{\infty}$.

  4. If $\inf I = 0$, from (1) we know $\lim_{p\to 0+}\|f\|_p$ exists.

Question: How $\Vert f \Vert_p$ behaves when $p$ is the endpoint of $I$?

To be more precise, assume $I$ is nonempty. Let $p_0 = \sup I$ (only right endpoint is considered for simplicity):

  • $p_0\in I$ or $p_0 \not\in I$?
  • Can $I$ be a singleton?
  • What if $p_0 = +\infty$?
  • What if $0<\mu(X) < +\infty$?

Please also notify me if there is any mistake above.

Best Answer

The question: $p_0 \in I$ or not ...
$X = [0,1/2]$ with Lebesgue measure. $$ f(x) = \frac{1}{x},\qquad I = (0,1), \\ f(x) = \frac{1}{x\log^2(1/x)},\qquad I = (0,1]. $$

You can get similar examples where $I=(1,+\infty)$ or $I=[1,+\infty)$. Add two examples (on disjoint subsets of the measure space), one example with $I = (0,1]$ and one with $I = [1,+\infty)$, to get an example with $I = \{1\}$.

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