$\||f|^{2}f\|_{H^{s}}\leq C \|f\|^{2}_{L^{\infty}} \|f\|_{H^{s}}$ for $s>0, f\in H^{s}(\mathbb R)$? We let $H^{s}(\mathbb R^{n}), (s\in \mathbb R)$  the Sobolev spaces.
It is well known that: the space $H^{s}(\mathbb R^{n})$ is an algebra with respect to pointwise multiplications, for $s>n/2.$

My Question: Can we expect $\||f|^{2}f\|_{H^{s}}\leq C \|f\|^{2}_{L^{\infty}} \|f\|_{H^{s}}$ for $s>0, f\in H^{s}(\mathbb R)$  (where $C$ is some constant)? If Yes, how to prove it?

Thanks,
 A: For $0<s<1$ this is true in all dimensions $n$, and follows by writing the norm as in integral of divided differences [Leoni, A first course in Sobolev spaces, 14.8]
$$
\|f\|_{H^s}^2 \approx \|f\|_{L^2}^2+\iint\frac{|f(x)-f(y)|^2}{|x-y|^{n+2s}}\,dx\,dy \tag{1}
$$
Indeed, (1) shows at once that for any Lipschitz function $\varphi$ fixing $0$ we have  $$\|\varphi\circ f\|_{H^s} \le C(\operatorname{Lip} \varphi)\| f\|_{H^s}\tag{2}$$
The function $\varphi(u)= |u|^2u$ is Lipschitz on the range of $f$.
For $s=1$ the statement also holds in all dimensions $n$, and is easy to prove from the definition. 

In higher smoothness orders there are issues in dimensions $n>1$. As an illustration: the third derivative of $f^3$ involves $(f')^3$, so we are looking to bound $\int |f'|^6$ in terms of $H^3$ norm of $f$. The Sobolev embedding gives such a bound in low dimensions only; in higher dimensions it fails, and the boundedness of $f$ is of no help here. 
On this topic, see Superposition in homogeneous and vector valued Sobolev spaces by Gérard Bourdaud  and the survey Composition Operators on Function Spaces with Fractional Order of Smoothness by Bourdaud and Sickel.

But in $n=1$ there is no such issue: all lower derivatives are neatly controlled by higher order derivatives. For integer exponents $s$ direct computation works: the $k$th derivative of $|f|^2 f$ splits into a bunch of terms like $D^a f\ D^b f\ D^cf$ with $a+b+c=k$, and the $L^2$ norms  of all of these are controlled by the $H^k$ norm of $f$. 

For fractional exponents $s>1$ one can try to differentiate $\lfloor s\rfloor$ times (again, expanding by the product rule) and then use the result for $s\in (0,1)$. However, this looks messy and I'm not quite sure this works out. It's better to use a characterization similar to (1) in which the first difference $|f(x)-f(y)|$ is replaced by higher order difference $|\Delta^k f|$. Then you can work with $|\Delta^k (|f|^2f)|$ algebraically, and  stick the estimates into the integral formula. A reference for the higher-order difference characterization of Sobolev spaces   is Integral Representations of Functions and Imbedding Theorems by Besov, Il'in, Nikolski.
