Convergence of $L^{p}$ norm as function of $p$.

Question

One can prove that in any measure space, if $\{p\in[1,\infty): \|f\|_{L^{p}}<\infty\}$ contains of two points, then this set is actually a connected subset of $\bf{R}$. Assume that it is of the form $[1,b]$, where $b<\infty$, can we have $\lim_{p\rightarrow 1^{+}}\|f\|_{L^{p}}\ne\|f\|_{L^{1}}$? I can only prove that whenever $f\in L^{\infty}$, then the limit is always the case, that's the reason I make the assumption that $b<\infty$.

Answer 1

Let $f \in L^1 \cap L^b$, with $f$ not a.e. $0$. Define a function $\phi: [1,b] \to (0,\infty)$ by sending $q \mapsto \int |f|^q d\mu$.

We will show that $\phi$ is continuous. Let $q_n$ be a sequence in $[1,b]$ such that $q_n \to q \in [1,b]$. Then $\;|f|^{q_n} \to |f|^q$. Moreover, for all $n$ it is true that $|f|^{q_n} \leq \max\{|f|,|f|^b\} \leq |f|+|f|^b \in L^1$. Hence by the dominated convergence theorem $\int |f|^{q_n} d\mu \to \int |f|^q d\mu$. This shows that $\phi$ is continuous.

Since $\phi$ is continuous, it follows that the map $q \mapsto \phi(q)^{1/q}$ is also continuous, since this is the composition of two continuous maps: $q \mapsto (f(q),\frac{1}{q})$, and $(x,y) \mapsto x^y$. For clarity, the first map is continuous from $[1,b] \to (0,\infty)^2$ and the second map is continuous from $(0,\infty)^2 \to \Bbb R$.

But $\phi(q)^{1/q}=\|f\|_q$, so that $q \mapsto \|f\|_q$ is continuous on $[1,b]$. In particular, we must have that $\lim_{q \to 1^+} \|f\|_q = \|f\|_1$.