Let $(X, \mathfrak{M}, \mu)$ is a measure space. Let $f:X \rightarrow \mathbb{C}$ be a measurable function. Prove that the set $\{1 \le p \le \infty \, | \, f \in L^p(X)\}$ is connected.
In other words prove that if $f \in L^p(X) \cap L^q(X)$ with$ 1\le p < q \le \infty$ and if $p\le r \le q$, then $f \in L^r(X)$.
I've tried to prove this using Holder's Inequality, here's my attempt:
Since $r \in [p,q]$, let $\lambda$ be such that $(1-\lambda)p+\lambda q = r.$ We use the fact that $f \in L^r(X)$ iff $|f|^r \in L^1(X)$. Let $m = \frac{1}{1-\lambda}$ and $n=\frac{1}{\lambda}$ so that $m$ and $n$ are conjugates.
Then, \begin{eqnarray*}
\int_{X} |f|^r \, d\mu &=& \int_{X} |f|^{(1-\lambda)p+\lambda q} \, d\mu \\
&=& \int_X |f|^{(1-\lambda)p}|f|^{\lambda q} d\mu \\
&\stackrel{\text{H$\ddot{o}$lder}}{\leq}& \left(\int_X |f|^p \, d\mu\right)^{(1-\lambda)}\left(\int_X |f|^q \, d\mu \right)^\lambda \, \\
&<& \infty
\end{eqnarray*}
Since surely the RHS is finite since $f \in L^p(X) \cap L^q(X)$.
Is this correct? Is this the intended method of proof?
Best Answer
Alternative:
$$\int\left|f\right|^{r}d\mu=\int_{\left\{ \left|f\right|\leq1\right\} }\left|f\right|^{r}d\mu+\int_{\left\{ \left|f\right|>1\right\} }\left|f\right|^{r}d\mu\leq\int_{\left\{ \left|f\right|\leq1\right\} }\left|f\right|^{p}d\mu+\int_{\left\{ \left|f\right|>1\right\} }\left|f\right|^{q}d\mu$$$$\leq\int\left|f\right|^{p}d\mu+\int\left|f\right|^{q}d\mu<\infty$$