Can you always bound the derivative of a Sobolev norm by a higher Sobolev norm in $\mathbb R$ and $\mathbb R^n$? Background: While reading a paper written by a mathematician, I noticed that they frequently "bounded the derivative of a function" by a Sobolev norm of one higher derivative. This was done when working against a subset of $\mathbb R$. Does this essentially fall out of the definition of the one-dimensional Sobolev norms? Does a similar result hold in $\mathbb R^n$?
Claim: Given any positive integer $s$ and $u\in H^s(\mathbb R)$, we have
$$
\|\partial_x u\|_{H^s(\mathbb R)} \le \|u\|_{H^{s+1}(\mathbb R)}.
$$
Proof: By the definition of the one-dimensional Sobolev norm,
$$
\|u\|_{H^s(\mathbb R)}^2 = \sum_{j=0}^s \|\partial_x^j u\|_{L^2(\mathbb R)}^2.
$$
Upon setting $\ell = j+1$, we discover
\begin{align*}
\|\partial_x u\|_{H^s(\mathbb R)}^2 &= 
\sum_{j=0}^s \|\partial_x^{j+1} u\|_{L^2(\mathbb R)}^2 \\&=
\sum_{\ell=1}^{s+1}\|\partial_x^{\ell} u\|_{L^2(\mathbb R)}^2 \\&\le \sum_{j=0}^{s+1} \|\partial_x^{j} u\|_{L^2(\mathbb R)}^2 \\&=
\| u\|_{H^{s+1}(\mathbb R)}^2.
\end{align*}
I would "assume" that something like this would be true for subsets of $\mathbb R^n$.
Claim 2: Given any positive integer $s$, multi-index $\beta$ where $|\beta|\leq s$, and $u\in H^s(\mathbb R^n)$, we have
$$
\|D^{\beta} u\|_{H^s(\mathbb R^n)} \le \|u\|_{H^{s+?}(\mathbb R^n)}.
$$
Proof 2: For every multi-index $\alpha$ where $|\alpha|\leq s$, we have by the definition of the multi-dimensional Sobolev norms,
$$
\| u \|_{H^s(\mathbb R^n)}^2 = \sum_{|\alpha | \leq s} \left \| D^{\alpha}u \right \|_{L^2(\mathbb R^n)}^2 
$$
Upon setting $\gamma = \alpha+\beta$, we discover
\begin{align*}
\|D^{\beta} u\|_{H^s(\mathbb R^n)}^2 &= 
\sum_{|\alpha | \leq s} \left \| D^{\alpha}D^{\beta} u \right \|_{L^2(\mathbb R^n)}^2 \\&= 
\sum_{|\gamma -\beta | \leq s} \left \| D^{\gamma} u \right \|_{L^2(\mathbb R^n)}^2 \\&\leq 
\sum_{|\gamma| \leq s} \left \| D^{\gamma} u \right \|_{L^2(\mathbb R^n)}^2 \\&= \|u\|_{H^{s+\beta}(\mathbb R^n)}.
\end{align*}
I don't think what I wrote in claim 2 is correct. To me, both of these results look like variations of the Poincaré inequality. Is there a theorem from Sobolev space theory which gives these results?
 A: I copy what you wrote to number the lines- \begin{align*}
\|D^{\beta} u\|_{H^s(\mathbb R^n)}^2 &= 
\sum_{|\alpha | \leq s} \left \| D^{\alpha}D^{\beta} u \right \|_{L^2(\mathbb R^n)}^2 \tag1\\&= 
\sum_{|\gamma -\beta | \leq s} \left \| D^{\gamma} u \right \|_{L^2(\mathbb R^n)}^2\tag2 \\&\leq \tag3
\sum_{|\gamma| \leq s} \left \| D^{\gamma} u \right \|_{L^2(\mathbb R^n)}^2 \\&= \|u\|_{H^{s+\beta}(\mathbb R^n)}. \tag4
\end{align*}
I'm not a fan of how  (2) is written, because the fact that $\gamma-\beta\ge 0$ (by which I mean a component-wise inequality) is hidden in the definition of a multi-index. (2) to (3) is wrong, consider the 1D case with $\beta=100$ and $s=1$. Then $|\gamma-100|\le 1$ and $|\gamma|\le 1$ describe disjoint sets of indices $\gamma$. Furthermore, as there is no mention of $\beta$ in (3), there is no way you can get that (3) is equal to (4), which does involve $\beta$.
I would have not bothered to write a line (2), and simply claim
\begin{align*}
\|D^{\beta} u\|_{H^s(\mathbb R^n)}^2 &= 
\sum_{|\alpha | \leq s} \left \| D^{\alpha}D^{\beta} u \right \|_{L^2(\mathbb R^n)}^2 \\&\leq 
\sum_{|\gamma| \leq s+|\beta|} \left \| D^{\gamma} u \right \|_{L^2(\mathbb R^n)}^2 \\&= \|u\|^2_{H^{s+|\beta|}(\mathbb R^n)}.
\end{align*}
where the inequality is justified since a derivative of the form
$$ D^\alpha D^\beta, |\alpha|\le s$$
is in particular, a derivative of order at most $s+|\beta|$ (and there are no repeats).
PS as mentioned in the comments Poincaré is the reverse inequality, involving the control of many terms by a few high order terms and is wrong without extra conditions (e.g. mean zero, vanishing on the boundary, etc.)
PPS if a paper writes $H^s$, then there is a good chance they are talking about $s\in\mathbb R$, rather than merely $s\in\mathbb Z_{\ge0}$. Here the space is defined using the Fourier transform (and agrees with the integer case). The proof is similarly easy, this time coming from a polynomial inequality $|\xi^\beta|\le (1+|\xi|^2)^{|\beta|/2}
$.
