If $\,-u''+u=f$, then $\,\|u\|_{L^s}+\|u'\|_{L^q}+\|u''\|_{L^p}\le c\|f\|_{L^p}$

Question

Let $u\in \mathcal{S}(\mathbb{R})$ (Schwartz space) be a solution of the equation $$ -u''+u=f, \quad \text{where}\,\,\,f \in \mathcal{S}(\mathbb{R}). $$ Show that for all $1\leq p,q,s \leq \infty$, there exists a $c>0$, such that $$ \|u\|_{L^s(\mathbb{R})}+\|u'\|_{L^q(\mathbb{R})}+\|u''\|_{L^p(\mathbb{R})} \le c\|f\|_{L^p(\mathbb{R})}. $$ This reminds me of the generalized Hölder inequality $$\int fgh \leq ||f||_{L^s(\mathbb{R})}||g||_{L^q(\mathbb{R})}||h||_{L^p(\mathbb{R})}$$ where $\frac1s+\frac1q+\frac1p=1$. How can I write $u$ as an integral $I(f)$?

Answer 1

Using Fourier transform and inverse Fourier transform we obtain $$ u(x)=-\frac{1}{2}\int_{-\infty}^\infty \mathrm{e}^{-|x-y|}f(y)\,dy, \\ u'(x)=-\frac{1}{2}\int_{-\infty}^\infty \mathrm{e}^{-|x-y|}\text{sgn}(x-y)\,f(y)\,dy, $$ Using Young's inequality we obtain $$ \|u\|_q\le c_0(q)\|f\|_p,\quad \|u’\|_s\le c_1(s)\|f\|_p $$ where $$ c_0(q)=\frac{1}{2}\big\|\mathrm{e}^{-|x|}\big\|_{q'}, \quad c_1(s)=\frac{1}{2}\big\|\mathrm{e}^{-|x|}\big\|_{s'} $$ and $q',s'$ such that $$ \frac{1}{p}+\frac{1}{q'}=\frac{1}{q}+1, \quad \frac{1}{p}+\frac{1}{s'}=\frac{1}{s}+1. $$ and $$ \|u''\|_p=\|u-f\|_p\le \|f\|_p+c_0(p)\|f\|_p $$ Altogether $$ \|u\|_q+\|u'\|_s+\|u''\|_p\le\big(1+c_0(p)+c_0(q)+c_1(s)\big)\|f\|_p. $$