Show that $f \in L^2(\mathbb R)$ Let $1\le p < 2 < q \le \infty$. Show that if $f\in L^p(\mathbb R)\cap L^q(\mathbb R)$, then $f\in L^2(\mathbb R)$.The hint is to use Holder and $a=a-b+b$.
I tried to start of with:
$\int_R |f|^2 \le (\int_R|f|^p)^{\frac{1}{p}}\cdot (\int_R |f|^q)^{\frac{p-1}{p}}$ and then to try to work on the second term to show that it is finite but to no result. Any ideas on how to do it?
 A: Assume $q<\infty$. Note that you can express 2 as a convex linear combination of $p$ and $q$, that is
$$
2=sp+tq,\quad s=\frac{q-2}{q-p}, \;t=\frac{2-p}{q-p}
$$
And we have $s+t=1$, $0<s,t<1$. If we set $r=s^{-1}$ and $r'=t^{-1}$, then $r$ and $r'$ are conjugate exponents such that $|f|^{sp}\in L^r(\mathbb{R})$ and $|f|^{tq}\in L^{r'}(\mathbb{R})$ since $f\in L^p(\mathbb{R})\cap L^q(\mathbb{R})$. Therefore, Holder's inequality yields
$$
\int |f|^2 dx=\int |f|^{sp}|f|^{tq} dx\leq \bigl(\int |f|^{rsp}dx\bigr)^\frac{1}{r}\bigl(\int |f|^{r'tq}dx\bigr)^\frac{1}{r'}=\bigl(\int |f|^{p}dx\bigr)^s\bigl(\int |f|^{q}dx\bigr)^t<\infty.
$$
(Note that this actually shows that $f\in L^a(\mathbb{R})$ for every $p<a<q$).
A: The $\ln$ function is concave. Therefore, for $x,y > 0$ and $0 \le t \le 1$, you have $(1-t)\ln x + t \ln y \le \ln((1-t)x+ty)$. Hence,
$$
          x^{1-t}y^{t} \le (1-t)x+ty.
$$
By taking a limit, the above continues to hold for $x \ge 0$ and $y \ge 0$.
Now replace $x$ by $|f|^{p}$ and $y$ by $|f|^{q}$ to obtain
$$
                      |f|^{(1-t)p+tq} \le (1-t)|f|^{p}+t|f|^{q}.
$$
Choose $t \in [0,1]$ so that $(1-t)p+tq=2$ and you get the result that $f \in L^{2}$, i.e., is square integrable.
