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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?

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    $\begingroup$ Looks ok to me. $\endgroup$ – Ivan Di Liberti Oct 17 '15 at 8:03
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    $\begingroup$ Related. See also log-convexity of $L^p$ norm. $\endgroup$ – A.Γ. Oct 17 '15 at 8:12
  • $\begingroup$ Note that $(1-\lambda)p+\lambda q = r$ is in trouble when $q=+\infty$. $\endgroup$ – A.Γ. Oct 17 '15 at 8:18
  • $\begingroup$ @A.G. Ah, indeed so. I shall look into a separate proof if $p$ is finite and $q$ is infinity. $\endgroup$ – Anthony Peter Oct 17 '15 at 8:20
  • $\begingroup$ @AnthonyPeter ...or consider the segment between $1/p$ and $1/q$ instead (see my link above to wiki). $\endgroup$ – A.Γ. Oct 17 '15 at 8:23
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Alternative:

$$\int\left|f\right|^{r}d\mu=\int_{\left\{ \left|f\right|\leq1\right\} }\left|f\right|^{r}d\mu+\int_{\left\{ \left|f\right|>1\right\} }\left|f\right|^{r}d\mu\leq\int_{\left\{ \left|f\right|\leq1\right\} }\left|f\right|^{p}d\mu+\int_{\left\{ \left|f\right|>1\right\} }\left|f\right|^{q}d\mu$$$$\leq\int\left|f\right|^{p}d\mu+\int\left|f\right|^{q}d\mu<\infty$$

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  • $\begingroup$ A slick answer indeed $\endgroup$ – Anthony Peter Oct 17 '15 at 8:12

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