Let $f \colon \mathbb{R} \to \mathbb{R}$ be a Lipschitz nonlinearity with $f(0) = 0$ and suppose $u \in \textrm{H}^s(\mathbb{R}) \cap \textrm{L}^\infty(\mathbb{R})$ for some $s \in [0, \tfrac{1}{2}]$. Given some threshold frequency $\xi_0$, define the operators $$ u_{\textrm{low}} := \mathscr{F}^{-1}(1_{\{ |\xi| < \xi_0 \}} \widehat{u}) \qquad \text{and} \qquad u_{\textrm{high}} := \mathscr{F}^{-1}(1_{\{ |\xi| \geq \xi_0 \}} \widehat{u}). $$ Do any of the estimates $$ \| f(u)_{\textrm{low}} \|_\infty \lesssim \| v \|_\infty \qquad \text{or} \qquad \| f(u)_{\textrm{high}} \|_\infty \lesssim \| v \|_\infty $$ hold, where $v$ is one of $u_\mathrm{low}$, $u_\mathrm{high}$ or $u$?
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$\begingroup$ Excluded $f(x) = ax + b$ functions. $\endgroup$– F. H.Commented Aug 2, 2019 at 17:26
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$\begingroup$ Why do you expect the estimate to be linear? if $f(u)=u^2$ for instance, a simple rescaling $u\mapsto tu$ shows that both estimates are impossible. $\endgroup$– Piero D'AnconaCommented Dec 31, 2019 at 7:18
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When $f$ is the constant function, the first claim can't possibly hold (I assume $f(u) is the composition?)
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$\begingroup$ $f(u)$ is the composition. Sorry, I only meant that $v$ is one of $u_\mathrm{low}$, $u_\mathrm{high}$, $u$, not all of them. Updated the question. $\endgroup$– F. H.Commented Aug 2, 2019 at 17:12
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$\begingroup$ When $f$ is constant, $f(u)_\textrm{low}=f$ is independent of $u$ and so cannot be bounded by anything that only depends on $u$. $\endgroup$ Commented Aug 2, 2019 at 17:14
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$\begingroup$ Somehow I read $f = \mathrm{identity}$ … Anyway, thanks for the observation—I never intended constant outer functions to be included as Lipschitz nonlinearities ;) Have updated the question once more. $\endgroup$– F. H.Commented Aug 2, 2019 at 17:17