Using Holder's inequality to prove interpolation property in $l^p$ sequence spaces I wish to prove that if $x \in l^q \cap l^q$ with $1 \leq q < r$ then for any $p$ with $q < p < r$ we have $||x||_{l^p} \leq ||x||_{l^q}^{q(r-p)/p(r-q)} ||x||_{l^r}^{r(p-q)/p(r-q)}$ using Holder's Inequality, which states that:
If $1\leq p,q \leq \infty$ and $\frac{1}{p}+\frac{1}{q} = 1$ then $\sum_{k=0}^\infty |x_ky_k| \leq ||x||_{l^p}||x||_{l^q}$.
So far, I have shown that $x \in l^p$ and noted that $q(r-p)/p(r-q) + r(p-q)/p(r-q) = 1$, so tried to use Holder's Theorem with $p(r-q)/q(r-p)$ and $p(r-q)/r(p-q)$, but then this gives the wrong norms. 
I'm really unsure as to how to plug this in to Holder's Theorem to give the right exponentials, any advice would be great!
 A: You are on the right track. You have to choose $a$ and $b$ with $a+b=p$ in such a way that you can apply Holder's inequality in
$$
\|x\|_p^p=\sum |x_i|^p=\sum |x_i|^a|x_i|^b
$$
with exponents $l$ and $m$ such that
$$
\frac{1}{l}+\frac{1}{m}=1
$$
(to be able to use the inequality) and, moreover, $al=q$, $bm=r$ (for the $q$-norm and the $r$-norm to show up). You solve for $a$, $b$, $l$ and $m$ satisfying the four conditions and you get
$$
a=\frac{q(r-p)}{r-q};\quad b=\frac{r(p-q)}{r-q};\quad l=\frac{r-q}{r-p};\quad m=\frac{r-q}{p-q}
$$
You then use Holder's inequality:
$$
\|x\|_p^p=\sum |x_i|^a|x_i|^b\le (\sum |x_i|^q)^{\frac{r-p}{r-q}}(\sum |x_i|^r)^{\frac{p-q}{r-q}}=\|x\|_q^{\frac{q(r-p)}{r-q}} \|x\|_r^{\frac{r(p-q)}{r-q}}
$$
as desired.
A: There is this following version of Hölder's inequality:
Take $p \in (q,r)$ and consider $\frac{1}{p} = \frac{\vartheta}{q} + \frac{1-\vartheta}{r}$ where $\vartheta \in (0,1)$. Then you have
$$
\|f\|_p \leq \|f\|_r^{1-\vartheta} \|f\|_q^{\vartheta}.
$$
The proof goes a bit likes this: First set $\vartheta := \frac{r-p}{r-q} \in (0,1)$ from which we get $p = \vartheta q + (1-\vartheta)r$. Using this we get
$$
\|f\|^p_p = \||f|^p\|_1 = \| |f|^{\vartheta q} |f|^{(1-\vartheta)r}\|_1 \leq \||f|^{\vartheta q}\|_{\frac{1}{\vartheta}} \| \||f|^{(1-\vartheta)r}\|_{\frac{1}{1-\vartheta}} = 
\|f\|_{q}^{q\vartheta} \| \|f\|_{r}^{r(1-\vartheta)},
$$
where the inequality is a use of Hoelder. Now, devide by $p$ and arrive at
$$
\|f\|^p \leq 
\|f\|_{q}^{\frac{q}{p}\vartheta} \| \|f\|_{r}^{\frac{r}{p}(1-\vartheta)}.
$$
To get to your result, just notice that by setting $\vartheta := \frac{r-p}{r-q} \in (0,1)$ you get your result.
