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I am reading Aspects of Sobolev-Type Inequalities by professor Laurent Saloff-Coste, where I found a claim on page 66 claiming the following:

Denote the following inequality as $S_{r,s}^{\theta}$: $\forall f \in C^{\infty}_{0}(M)$, $||f||_r \leq (C||\nabla f||_p)^{\theta}||f||_s^{1-\theta}$, where M is a complete non-compact Riemannian manifold and the parameters satisfy $0<\theta\leq 1$ and $0 < s <r\leq\infty$.

Now the claim is: $\forall 0 < s < r < \infty$, $\forall 0<s'<r'<\infty$, such that $r' \leq r$ and $s'\leq s$, and $\frac{1}{r} = \frac{1-\theta}{s}$ and $\frac{1}{r'} = \frac{1-\theta'}{s'}$, we have that $S_{r,s}^{\theta}$ implies $S_{r',s'}^{\theta'}$ by Holder inequality.

My question is how can we prove this claim. This should be elementary but from simple applications of Holder inequality I can only see the claim when $r'\leq r$ and $s' = s$. How can I go to the case where $s'\leq s$. The direction of inequality in $S_{r,s}^{\theta}$ seems to go against Holder's inequality.

Thank you very much for the help.

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It is indeed elementary with some slight maneuvering:

Since $s' < r' \leq r$, there exists $\alpha \in (0,1]$ such that \begin{equation}\|f\|_{r'} \lesssim \|f\|_{s'}^{1-\alpha} \, \|f\|_r^\alpha.\label{1}\tag{1}\end{equation} Hence (by the $S^\theta_{r,s}$ assumption) \begin{equation}\|f\|_{r'} \lesssim \|f\|_{s'}^{1-\alpha} \, \|f\|_s^{\alpha(1-\theta)}\,\|\nabla f\|_p^{\alpha\theta}. \label{2}\tag{2}\end{equation}

Now consider the cases (a) $s' < s \leq r'$ and (b) $r' \leq s < r$. For (a), you obtain a bound on $\|f\|_s$ in terms of $\|f\|_{s'}$ and $\|f\|_{r'}$ just like \eqref{1} which you can insert in \eqref{2} to get $S^{\theta'}_{r',s'}$ for some appropriate $\theta'$ to be calculated. For (b), the same ansatz gives a bound on $\|f\|_s$ in terms of $\|f\|_{r'}$ and $\|f\|_r$. Using assumption $S^\theta_{r,s}$, you get $S^\tau_{s,r'}$ for some $\tau \in [0,1)$ which can then be inserted into \eqref{1} to yield $S^{\theta'}_{r',s'}$.

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