Let $\Omega$ be a domain in $\mathbb{R}$. Let $\phi_{1}, \phi_{2} \in H^{1}_{0, \sigma}, \phi_{3} \in L^{r}$, where $n \leq r \leq \infty$. Also $H^{1}_{0, \sigma}$ is the closure of $C^{\infty}_{0, \sigma}$ with respect to the norm:
$||\phi||_{H^{1}} = ||\phi||_{L^{2}} + ||\nabla \phi||_{L^{2}}$
$C^{\infty}_{0, \sigma}$ is the set of all $C^{\infty}_{0}$ vector functions $\phi = (\phi_{1}, ..., \phi_{n})$ with support in $\Omega$ such that $\nabla \cdot \phi = 0$.
Then $||\phi_{1} \phi_{3}||_{L^{2}} \leq M ||\nabla \phi_{1}||^{n/r}_{L^{2}} ||\phi_{1}||^{1 - n/r}_{L^{2}} ||\phi_{3}||^{}_{L^{r}}$.
The proof of this statement begins with the below use of Holder`s Inequality, but I cannot see how this immediately follows.
"By Holder`s Inequality $M||\phi_{1} \phi_{3}||_{L^{2}} \leq M ||\phi_{1}||^{n/r}_{L^{m}} ||\phi_{1}||^{1 - n/r}_{L^{2}} ||\phi_{3}||^{}_{L^{r}}$, where $\frac{1}{m} = \frac{1}{2} - \frac{1}{n}$"
I am familiar with the original statement of Holder`s Inequality, but I have before seen this technique where 1 norm is split into 3. Could someone please explain at least the first few steps towards showing this statement is true.
The general version of Hölder's Inequality states that $$ \|f_1\ldots f_k\|_p\leq \|f_1\|_{p_1}\ldots\|f_k\|_{p_k} $$ whenever $$ \frac{1}{p}=\frac{1}{p_1}+\cdots+\frac{1}{p_k}. $$
In fact, there is an interpolated form of the inequality which is more convenient to apply to the question at hand (also at the same link) stating $$ \bigl\||f_1|^{\theta_1}\ldots |f_k|^{\theta_k}\bigr\|_p\leq \|f_1\|^{\theta_1}_{p_1}\ldots \|f_k\|^{\theta_k}_{p_k}, $$ where $$ \frac{1}{p}=\frac{\theta_1}{p_1}+\cdots+\frac{\theta_k}{p_k}. $$ Taking $k=3$ and $p=2$ and $$ (\theta_1,\theta_2,\theta_3)=\bigl(\tfrac{n}{r},1-\tfrac{n}{r},1\bigr),\quad (p_1,p_2,p_3)=(m,2,r),\quad (f_1,f_2,f_3)=(\phi_1,\phi_1,\phi_3) $$ yields the desired inequality. Note that this is a consistent choice of parameters, since $$ \frac{1}{2}=\frac{n/r}{m}+\frac{1-n/r}{2}+\frac{1}{r} $$ is equivalent to $$ \frac{1}{m}=\frac{1}{2}-\frac{1}{n}. $$