Problem: Let $d\geq 2,\,1<p<d$, and $1\leq q<r<\frac{dp}{d-p}$. Here $d$ represents the space dimension. For some $\theta\in (0,1)$ and some constant $C>0$ we have
$$\|u\| _{L^r(\mathbb R^d)}\leq C \|u\| _{L^q(\mathbb R^d)}^{1-\theta}\|\nabla u\| _{L^p(\mathbb R^d)}^{\theta}\quad\text{for all }u\in\mathcal C^{\infty}_c(\mathbb R^d).$$
$\textbf{a)}$ Use scaling to find the $\theta$.
$\textbf{b)}$ Prove the inequality and state on which parameters is $C$ depending.
$\textit{Hint:}$ Do an interpretation of $L^r$ in terms of $L^q$ and $L^{\frac{dp}{d-p}}$, and then apply Sobolev inequality.
My Attempt: Throughtout the proof $u\in\mathcal C_c^\infty(\mathbb R^d)\setminus\{0\}$. For the scaling argument, fix $\lambda>0$ and define $u_\lambda(x)=u(\lambda x)$. Now we compute $$\int_{\mathbb R^d}\vert u_\lambda\vert^r\,dx=\frac{1}{\lambda^d}\int_{\mathbb R^d}\vert u(y)\vert^r\,dy,$$ $$\int_{\mathbb R^d}\vert u_\lambda\vert^q\,dx=\frac{1}{\lambda^d}\int_{\mathbb R^d}\vert u(y)\vert^q\,dy,$$ $$\int_{\mathbb R^d}\vert \nabla u_\lambda\vert^p\,dx=\frac{\lambda^p}{\lambda^d}\int_{\mathbb R^d}\vert \nabla u(y)\vert^p\,dy.$$ Putting the above back into the proposed inequality we get that $$\frac{1}{\lambda^{d/r}}\|u\|_{L^r(\mathbb R^d)}\leq C\frac{1}{\lambda^{d(1-\theta)/q}}\|u\|_{L^q(\mathbb R^d)}\lambda^{(1-d/p)\theta}\|\nabla u\|_{L^p(\mathbb R^d)}.$$ Hence $$\|u\|_{L^r(\mathbb R^d)}\leq C\lambda^{(1-d/p)\theta+(1-\theta)d/p-1/r}\|u\|_{L^q(\mathbb R^d)}\|\nabla u\|_{L^p(\mathbb R^d)}.$$ If $(1-d/p)\theta+(1-\theta)d/p-1/r\ne0$, then we can upon sending $\lambda$ to either $0$ or $\infty$ obtain a contradiction. Therefore we must have $$\left(1-\frac{d}p\right)\theta+(1-\theta)\frac{d}p-\frac1r=0.$$ Now we prove the inequality does in fact hold. First we prove the following lemma.
Lemma: Let $1\leq p<r<q\leq\infty$ and assume that $f\in L^{p}(\mathbb R^d)\cap L^{q}(\mathbb R^d).$ Prove that
$$\|f\|_{r}\leq \|f\|_{p}^{\theta}\|f\|_{q}^{1-\theta}$$
where $\theta\in(0,1)$ such that $\frac{1}{r}=\frac{\theta}{p}+\frac{1-\theta}{q}.$ In particular $f\in L^{r}(\mathbb R^d)$.
First we work out the case $1<p<r<q<\infty.$ Put
$$\theta=\frac{p(q-r)}{r(q-p)}.$$
Since $rq>pq>0$ by hypothesis, we have that $r(q-p)>p(q-r)>0$ and dividing each side of the inequality by $r(q-p)$, we see that $0<\theta<1.$
By construction, we have $\frac{pq}{r}-p=\theta(q-p),$ so that
$$\frac{1}{r}=\frac{\theta}{p}+\frac{1-\theta}{q},\quad\text{whence}\quad1=\frac{1}{p/r\theta}+\frac{1}{q/r(1-\theta)}.$$
Since $p/r\theta,q/r(1-\theta)>1,$ we can use Holder's inequality to obtain
\begin{equation*}\begin{split}
\|f^{r\theta}f^{r(1-\theta)}\|_{1} &= \int_{\mathbb R^d}\vert f(x)\vert^{r\theta}\vert f(x)\vert^{r(1-\theta)}\,dx \\
&\leq \left(\int_{\mathbb R^d} \left(\vert f(x)\vert^{r\theta}\right)^{\frac{p}{r\theta}}\,dx \right)^{\frac{r\theta}{p}} \left( \int_{\mathbb R^d} \left(\vert f(x)\vert^{r(1-\theta)}\right)^{\frac{q}{r(1-\theta)}}
\,dx \right)^{\frac{r(1-\theta)}{q}}\\
&= \|f\|_{p}^{r\theta}\|f\|_{q}^{r(1-\theta)}.
\end{split}\end{equation*}
and taking $r$th roots on both sides, we get that $\|f\|_{r}\leq \|f\|_{p}^{\theta}\|f\|^{1-\theta}_{q}.$
Finally, the hypothesis that $f\in L^{p}(\mathbb R^d)\cap L^{q}(\mathbb R^d)$ implies that $f\in L^{r}(\mathbb R^d).$ Now we proceed to handle the case $1=p<r<q=\infty.$ We may assume that $\vert f\vert\leq\|f\|_{\infty}$ for all $x\in\mathbb R^d.$ Putting $\theta=1/r,$ we see that
\begin{equation*}\begin{split}
\int_{\mathbb R^d}\vert f(x)\vert ^{r\theta}\vert f(x)\vert^{r(1-\theta)}\,dx &= \int_{\mathbb R^d}\vert f(x)\vert\vert f(x)\vert^{r(1-\theta)}\,dx = \int_{\mathbb R^d}\vert f(x)\vert\vert f(x)\vert^{r-1}\,dx \\
& \leq \int_{\mathbb R^d} \vert f(x)\vert\|f\|_{\infty}^{r-1}\,dx \\
& \leq \|f\|_{1}\|f\|_{\infty}^{r-1},
\end{split}\end{equation*}
and taking $r$th roots we get that $\|f\|_{r}\leq\|f\|_{1}^{\theta}\|f\|_{\infty}^{1-\theta},$ whence $f\in L^{r}(\mathbb R^d).$ \newline
\noindent Next, we handle the case $1=p<r<q<\infty.$ Turns out that we can use Holder's inequality here. Put $\theta=\frac{q-r}{r(q-1)}.$
Then $0<\theta<1,$ and $\frac{1}{r}=\theta+\frac{1-\theta}{q},$ so that $1=\frac{1}{1/r\theta}+\frac{1}{q/r(1-\theta)},$ and since $1/r\theta,q/r(1-\theta)>1,$ we can use Holder's inequality to get
\begin{equation*}\begin{split}
\int_{\mathbb R^d}\vert f(x)\vert^{r\theta}\vert f(x)\vert^{r(1-\theta)}\,dx & \leq \left(\int_{\mathbb R^d}\left( \vert f(x)\vert^{r\theta} \right)^{\frac{1}{r\theta}}\,dx\right)^{r\theta}
\left(\int_{\mathbb R^d}\left(\vert f(x)\vert^{r(1-\theta)} \right)^{\frac{q}{r(1-\theta)}}\,dx\right)^{\frac{r(1-\theta)}{q}}\\
&=\|f\|_{1}^{r\theta}\|f\|_{q}^{r(1-\theta)},
\end{split}\end{equation*}
and taking $r$th roots we get that $\|f\|_{r}\leq\|f\|_{1}^{\theta}\|f\|_{q}^{1-\theta},$ and hence $f\in L^{r}(\mathbb R^d).$
Finally, we need to handle the case $1<p<r<q=\infty.$ Once again, we may assume that $\vert f(x)\vert\leq \|f\|_{\infty}$ for all $x\in\mathbb R^d.$ Put $\theta=p/r.$
Then $0<\theta<1, \frac{1}{r}=\tfrac{\theta}{p}$, and $r(1-\theta)>1$. With this in mind, we observe that
\begin{equation*}\begin{split}
\int_{\mathbb R^d}\vert f(x)\vert^{r\theta}\vert f(x)\vert^{r(1-\theta)}\,dx & \leq\int_{\mathbb R^d}\vert f(x)\vert^{p}\|f\|_{\infty}^{r(1-\theta}\,dx\\
&= \|f\|_{p}^{p}\|f\|_{\infty}^{r(1-\theta)}.
\end{split}\end{equation*}
Now taking the $r$th root in the above inequality, we see that $\|f\|_{r}\leq\|f\|_{p}^{\theta}\|f\|_{\infty}^{1-\theta},$ whence $f\in L^{r}(\mathbb R^d).$ And the proof is finally complete.
Back to business, we can apply the lemma with $1\leq q<r<\frac{dp}{d-p}$ and theta will be as found in the scaling argument, then the Gagliardo-Nirenberg-Sobolev inequality to obtain $$\|u\|_{L^r(\mathbb R^d)}\leq\|u\|_{L^q(\mathbb R^d)}^{1-\theta}\|u\|_{L^{\frac{dp}{d-p}}(\mathbb R^d)}^\theta\leq C\|u\|_{L^q(\mathbb R^d)}^{1-\theta}\|\nabla u\| _{L^p(\mathbb R^d)}^{\theta}.$$ Note that $C$ depends only on $d$ and $p$.
Do you agree with my proof above?
Thank you for your time and feedback.