$f \in L^p(\mathbb{R^d})\cap L^r(\mathbb{R^d})$ implies $f\in L^q(\mathbb{R^d})$ where $p<q<r$

Question

Prove that if $f \in L^p(\mathbb{R^d})\cap L^r(\mathbb{R^d})$ then $f\in L^q(\mathbb{R^d})$ where $p<q<r$. I have tried all the usual $L^p$ inequalities but dead end every time. I think I should use Cauchy or Holders to solve it but I can't find the way. Can anyone give me a hint so I can get started? Thank you in advance!

Answer 1

Suppose $$ \frac\alpha p+\frac{1-\alpha}r=\frac1q\implies\frac{\alpha q}p+\frac{(1-\alpha)q}r=1 $$ Then Hölder's Inequality says $$ \begin{align} \int_{\mathbb{R}^d}|f|^q\,\mathrm{d}x &=\int_{\mathbb{R}^d}|f|^{\alpha q}|f|^{(1-\alpha)q}\,\mathrm{d}x\\ &\le\left(\int_{\mathbb{R}^d}|f|^{\alpha q\large\frac p{\alpha q}}\,\mathrm{d}x\right)^{\large\frac{\alpha q}p} \left(\int_{\mathbb{R}^d}|f|^{(1-\alpha)q\large\frac r{(1-\alpha)q}}\,\mathrm{d}x\right)^{\large\frac{(1-\alpha)q}r}\\ &=\left[\left(\int_{\mathbb{R}^d}|f|^p\,\mathrm{d}x\right)^{\large\frac\alpha p} \left(\int_{\mathbb{R}^d}|f|^r\,\mathrm{d}x\right)^{\large\frac{1-\alpha}r}\right]^q\\ \end{align} $$ Therefore, $$ \|f\|_q\le\|f\|_p^\alpha\|f\|_r^{1-\alpha} $$

Answer 2

Hint: given $f$, define $A=\{ x : |f(x)| \geq 1 \}$. Then on $A$, $|f|^q \leq |f|^r$, and on $A^c$, $|f|^q \leq |f|^p$. Use this to get a bound on $|f|^q$ by a function which you know is in $L^1$. Conclude that $f \in L^q$ from there.

You can think about this proof as breaking up the obstruction to $f$ being in $L^q$ between its "singularities" (which lie on $A$) and its "tails" (which lie on $A^c$). It might be instructive to play with functions of the form $x^a 1_{[0,1]}(x) + x^b 1_{[1,\infty)}(x)$, where $b<a<0$, to get a feel for this.