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

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.

• By right endpoint of $I$ do you mean at $\infty$? – User203940 Oct 26 at 15:14
• @User203940 $I$ is the set of $p$ such that $\|f\|_p$ is finite. – Zhang Oct 26 at 15:15
• Ah I see, sorry I was skimming and not reading carefully. – User203940 Oct 26 at 15:16

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 $$I = \{1\}$$.
• Thank you. Are there similar examples when $p_0$ is $+\infty$? – Zhang Oct 27 at 0:49
• I see. If $p_0 = +\infty$, let $X = [0, 1]$, $f$ can be chosen to be $log(x)$ and constant $1$. – Zhang Oct 27 at 1:12