Inequality involving the Besov norm The $L^p$ norm of a function $f$ is given by
$$
\|f\|_{L^p(\mathbb{R}^d)}=\left(\int_{\mathbb{R}^d}|f(x)|^p\,dx\right)^{1/p}
$$
Let $\phi$ be a non-negative smooth function supported in $\{x\in\mathbb{R}^d:|x|\in(1/2,2)\}$ and such that for $x\neq 0$
$$
\sum_{j\in\mathbb{Z}}\phi(2^{-j}x)=1.
$$
Then the Besov norm of $f$ is defined by
$$
\|f\|_{B_{p,q}^s(\mathbb{R}^d)}=\left(\sum_{j\in\mathbb{Z}}2^{jsq}\|\mathcal{F}^{-1}\left(\phi(2^{-j}\cdot)\mathcal{F}f \right)\|^q_{L^p(\mathbb{R}^d)}\right)^{1/q}.
$$
I would like to use the inequality of the form
\begin{equation}
\|f\|_{B_{p,1}^{d/p}(\mathbb{R}^d)}\le C \|f\|^{1-d/p}_{L^p(\mathbb{R}^d)}\|f'\|^{d/{p}}_{L^p(\mathbb{R}^d)},\qquad (*)
\end{equation}
with some constant $C>0$ independent of $f$.
My $\textbf{question}$ is: is the inequality $(*)$ true and if so, is there any reference to read about it? It looks like some kind of interpolation inequality, but I'm not too familiar with the interpolation theory of function spaces.
I'd appreciate any help!
 A: Yes, it is an interpolation inequality.
First remark: $B^{d/p}_{p,1}$ says more or less that $d/p$ derivatives are bounded in $L^p$ norm. Since you are trying to control it using both $0$ and $1$ derivatives, you need at least $d/p ∈ [0,1]$.

*

*When $d/p\in (0,1)$, then the besov spaces are exactly the real interpolation spaces between $L^p$ and $W^{1,p}$ spaces (see e.g. Lunardi, Interpolation Theory section 1.1.1),
$$
(L^p,W^{1,p})_{d/p,1} = B^{d/p}_{p,1}
$$
and from that we deduce
$$
\|f\|_{B^{d/p}_{p,1}} ≤ C\, \|f\|_{L^p}^{1-d/p} \, \|f'\|_{L^p}^{d/p}
$$
Remark that your exponents in the right of $(*)$ are wrong. The fact that there is an exponent $d/p$ on the norm of $f'$ is coherent with the fact that $B^{d/p}_{p,1}$ means "$d/p$ derivatives" and it can be checked by scaling.


*If $p=\infty$, then this is not true since $B^{0}_{\infty,1} ⊂ C^0$ (see for example Bahouri, Chemin, Danchin Fourier Analysis and Nonlinear Partial Differential Equations, Proposition 2.39) and so  $L^\infty\not\subset B^{0}_{\infty,1}$


*If $p=d$, this is not true since $B^{d/p}_{p,1} = B^{1}_{p,1}$ is strictly smaller than $W^{1,p} = F^1_{p,2} ⊃ F^1_{p,1} \supset B^{1}_{p,1}$ (Here, $F^s_{p,q}$ denotes the Triebel-Lizorkin spaces, see for example the book of Triebel, Theory of Function Spaces II)
