If $f \in L^{p_1}$ and $f \in L^{p_2}$ with $1 \le p_1 \lt p_2 \lt \infty$, then $f \in L^{p}$ for all $p$ such that $p_1 \leq p \leq p_2$.

Question

Let $(X, \Sigma, \mu)$ be a measure space. If $f \in L^{p_1}$ and $f \in L^{p_2}$ with $1 \le p_1 \lt p_2 \lt \infty$, then $f \in L^{p}$ for all $p$ such that $p_1 \leq p \leq p_2$.

My attempt

Let $E = \{ x \in X : |f(x)| \leq 1\}$. Thus

Since $|f(x)|^p \leq |f(x)|^{p_1} $, for all $x \in E$, we have:

$$\int_{E}{|f|^p}d\mu \leq\int_{E}{|f|^{p_1}}d\mu \lt +\infty $$

For all $x \in X-E$, we have $|f(x)|^p \leq |f(x)|^{p_2} $.

Then: $$\int_{X-E}{|f|^p}d\mu \leq\int_{X-E}{|f|^{p_2}}d\mu \lt +\infty $$

Thus $$\int_{X}{|f|^p}d\mu = \int_{E}{|f|^p}d\mu + \int_{X-E}{|f|^p}d\mu \lt \infty \Rightarrow f\in L^p$$

I'd like to know if my solution is correct and if there exists a better way to solve it.

Thanks in advance!

Answer 1

I'm not sure about better, but here is an alternate solution based on Holder's inequality: since $p_1 < p < p_2$ you have $\frac 1{p_2} < \frac 1p < \frac 1{p_1}$ so there exists a constant $0 < \alpha < 1$ with $\frac 1p = \frac{\alpha}{p_1} + \frac{1-\alpha}{p_2}$. Then $1 = \frac{p\alpha}{p_1} + \frac{p(1-\alpha)}{p_2}$ and by Holder's inequality $$ \int_X |f|^p\, d\mu = \int_X |f|^{p\alpha} |f|^{p(1-\alpha)} \, d\mu \le \left( \int_X |f|^{p_1} \, d\mu \right)^{p\alpha/p_1} \left( \int_X |f|^{p_2} \, d\mu \right)^{p(1-\alpha)/p_2}.$$