$f\in L_{p_1}\cap L_{p_2}$ implies $f\in L_{p}$ for all $p\in (p_1,p_2)$

Question

I've proved that $f\in L_{p_1}\cap L_{p_2}$ implies $f\in L_{p}$ for all $p\in (p_1,p_2),$ where $1\leq p_{1}<p_{2}<+\infty$ using the next hint of the book that I'm reading: there is $\alpha\in(0,1)$ such that $\frac{1}{p}=\frac{\alpha}{p_{1}}+\frac{(1-\alpha)}{p_{2}}.$

The proof follows by Hölder's inequality with $r=\frac{p_1}{\alpha p}, s=\frac{p_{2}}{p(1-\alpha)}\in(1,+\infty)$ such that $\frac{1}{r}+\frac{1}{s}=1.$

My question is: How can we ensure the existence of such $\alpha?$

Answer 1

For $a<b<c$, then $b=\dfrac{c-b}{c-a}a+\left(1-\dfrac{c-b}{c-a}\right)c$. Now plug in $b=\dfrac{1}{p}$, $a=\dfrac{1}{p_{2}}$, $c=\dfrac{1}{p_{1}}$, $\alpha=1-\dfrac{c-b}{c-a}=\dfrac{b-a}{c-a}$.