# Inequality in the unit ball of Sobolev Space $W^{1,1}(\mathbb{R})$

In the book Functional Analysis, Sobolev Spaces and Partial Differential Equations by Haim Brezis

The Sobolev space $$W^{1,p}(I)$$ is defined to be $$W^{1,p}(I)=\{u\in L^{p}(I);\exists g\in L^p(I)\text{ such that } \int_{I}{u\phi^{'}}=-\int_{I}g\phi\ \forall \phi\in C^{1}_c(I)\}$$ where $$C^{k}_c(I)$$ means continuous, k-time differentiable function on a compact subset of $$I.$$

For a extension operator $$P:W^{1,p}(I)\to W^{1,p}(\mathbb{R}),$$ satisfying the following properties: $$(i)\ Pu|_{I}=u,\ \forall u\in W^{1,p}(I),$$ $$(ii)\ ||Pu||_{L^{p}(\mathbb{R})}\le C||u||_{L^{p}(\mathbb{R})},\ \forall u\in W^{1,p}(I),$$ $$(iii)\ ||Pu||_{W^{1,p}(\mathbb{R})}\le C||u||_{W^{1,p}(\mathbb{R})},\ \forall u\in W^{1,p}(I),$$ where $$C$$ depends only on $$|I|\le\infty.$$

Let $$\mathcal{H}$$ be the unit ball in $$W^{1,1}(I).$$ Let $$P$$ be the extension operator as above and set $$\mathcal{F}=P(\mathcal{H}),$$ so that $$\mathcal{H}=\mathcal{F}|_{I}.$$ We prove that $$\mathcal{H}$$ has a compact closure in $$L^{q}(I)$$(for all $$1\le q<\infty$$) by applying Theorem 4.26(Kolmogorov-M.Riesz-Frechet).Now $$\mathcal{F}$$ is bounded in $$W^{1,1}(\mathbb{R});$$ therefore $$\mathcal{F}$$ is also bounded in $$L^{q}(\mathbb{R}),$$ since it is bounded both in $$L^{1}(\mathbb{R})$$ and in $$L^{\infty}(\mathbb{R}).$$ We now check the condition for Theorem 4.26 i.e. $$\lim_{h\to 0}||\tau_{h}f-f||_q=0,\text{ uniformly in }f\in\mathcal{F}.$$ where $$\tau_hf(x)=f(x+h).$$By Proposition 8.5 we have, for every $$f\in\mathcal{F}\subset W^{1,p}(\mathbb{R}),$$ $$||\tau_h f-f||_{L^1(\mathbb{R})}\le |h|||f^{'}||_{L^{1}(\mathbb{R})}\le C|h|,$$ thus $$||\tau_h f-f||^{q}_{L^{q}(\mathbb{R})}\le (2||f||_{L^{\infty}(\mathbb{R})})^{q-1}||\tau_h f-f||_{L^1(\mathbb{R})}\le C|h|,$$ and consequently $$||\tau_h f-f||_{L^q(\mathbb{R})}\le C|h|^{\frac{1}{q}}$$ where $$C$$ is independent of $$f.$$ My question is how do I obtain this inequality: $$||\tau_h f-f||^{q}_{L^{q}(\mathbb{R})}\le (2||f||_{L^{\infty}(\mathbb{R})})^{q-1}||\tau_h f-f||_{L^1(\mathbb{R})}$$

## 1 Answer

Short version: Hölder's inequality ...

Longer version: This is just the fact that since $$f(x) \leq \|f\|_{L^\infty}$$, $$\|f\|_{L^q}^q = \int_{\Bbb R} |f(x)|^q\,\mathrm d x \leq \|f\|_{L^\infty}^{q-1} \int_{\Bbb R} |f(x)|\,\mathrm d x$$ which is an endpoint of Hölder's inequality, and so $$\|\tau_h f -f\|_{L^q}^q \leq \|\tau_h f -f\|_{L^\infty}^{q-1}\,\|\tau_h f -f\|_{L^1} \leq (2\|f\|_{L^\infty})^{q-1}\,\|\tau_h f -f\|_{L^1}$$ where the last inequality follows by the triangle inequality and the fact that $$\|\tau_hf\|_{L^\infty} = \|f\|_{L^\infty}$$.