Endpoint trace bound and decay of H1 functions

[Bounded-interval trace bound] Let $u\in H^1(0,L)$ and fix $x\in [0,L]$. By [F1] applied on $(0,L)$ (and on $(c,d)$ for the local form), for every $y\in [0,L]$ $\mathrm{\mid }{u}^{\ast }(x)\mathrm{\mid }\le \mathrm{\mid }{u}^{\ast }(y)\mathrm{\mid }+\mid {\int }_y^x{u}^{\mathrm{\prime }}\mid \le \mathrm{\mid }{u}^{\ast }(y)\mathrm{\mid }+\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(0,L)}{L}^{1\mathrm{/}2},$ using [F2] for the last estimate. Integrating this inequality in $y$ over $(0,L)$ and using [F2] again, $L\mathrm{\mid }{u}^{\ast }(x)\mathrm{\mid }\le {\int }_0^L\mathrm{\mid }{u}^{\ast }\mathrm{\mid }+L^{3\mathrm{/}2}\mathrm{\parallel }{u}^{\mathrm{\prime }}\mathrm{\parallel }\le L^{1\mathrm{/}2}\mathrm{\parallel }u\mathrm{\parallel }+L^{3\mathrm{/}2}\mathrm{\parallel }{u}^{\mathrm{\prime }}\mathrm{\parallel },$ so $\mathrm{\mid }{u}^{\ast }(x)\mathrm{\mid }\le L^{-1\mathrm{/}2}\mathrm{\parallel }u\mathrm{\parallel }+L^{1\mathrm{/}2}\mathrm{\parallel }{u}^{\mathrm{\prime }}\mathrm{\parallel }$. Replacing $(0,L)$ by any subinterval $(c,d)$ with $x\in [c,d]$ gives the local form by the same computation. This is (a).

[Half-line pointwise and supremum bounds] Let $u\in H^1(0,\mathrm{\infty })$. For every $n$ the class lies in $H^1(0,n)$, and the representatives on overlapping intervals agree by uniqueness in [F1], so they glue to a continuous representative $u^{\ast }$ on $[0,\mathrm{\infty })$. Step 1.1 on the interval $(x,x+1)$, $x\ge 0$, gives $\mathrm{\mid }{u}^{\ast }(x)\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(x,x+1)}+\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(x,x+1)}$. Enlarging each interval to $(0,\mathrm{\infty })$ and using $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(x,x+1)}\le \mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(0,\mathrm{\infty })}$ and likewise for $u^{\mathrm{\prime }}$ gives ${\sup }_{x\ge 0}\mathrm{\mid }{u}^{\ast }(x)\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(0,\mathrm{\infty })}+\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(0,\mathrm{\infty })}$.

[Decay] We show ${\lim }_{x\to \mathrm{\infty }}{u}^{\ast }(x)=0$. First suppose $\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(0,\mathrm{\infty })}=:M>0$ and assume for contradiction that there are $x_n\to \mathrm{\infty }$ with $\mathrm{\mid }{u}^{\ast }(x_n)\mathrm{\mid }\ge \epsilon$ for some $\epsilon >0$. Put $\delta :=\min \{1,(\epsilon \mathrm{/}(2M){)}^2\}>0$; by [F1], for $t\in [x_n,x_n+\delta ]$ $\mathrm{\mid }{u}^{\ast }(t)\mathrm{\mid }\ge \mathrm{\mid }{u}^{\ast }(x_n)\mathrm{\mid }-\sqrt{\mathrm{\mid }t-x_n\mathrm{\mid }}M\ge \epsilon -\sqrt{\delta }M\ge \epsilon \mathrm{/}2.$ Passing to a subsequence, assume $x_{n+1}\ge x_n+\delta$ for all $n$; then the half-open intervals $E_n:=[x_n,x_n+\delta )$ are pairwise disjoint, have measure $\delta$, and satisfy $\mathrm{\mid }{u}^{\ast }{\mathrm{\mid }}^2\ge {\epsilon }^2\mathrm{/}4$ throughout. By [F3], ${\int }_0^{\mathrm{\infty }}\mathrm{\mid }u{\mathrm{\mid }}^2\ge {\int }_{{\bigcup }_n{E}_n}\mathrm{\mid }u{\mathrm{\mid }}^2={\sum }_n{\int }_{E_n}\mathrm{\mid }u{\mathrm{\mid }}^2\ge {\sum }_n\frac{{\epsilon }^2}{4}\delta =+\mathrm{\infty },$ contradicting $u\in L^2(0,\mathrm{\infty })$. If $M=0$, then [F1] gives $\mathrm{\mid }{u}^{\ast }(x)-u^{\ast }(y)\mathrm{\mid }\le 0$, so $u^{\ast }$ is a constant whose $L^2$ norm over the half-line is finite only if that constant is $0$. Thus $u^{\ast }(x)\to 0$.

[Real line] For $u\in H^1(\mathbb{R})$ restrict to $(0,\mathrm{\infty })$ and $(-\mathrm{\infty },0)$; the representatives glue at $0$ by uniqueness of the continuous representative. Steps 1.1 and 2.1 give the sup bound and uniform continuity, and step 3.1 applied to both half-lines gives $u^{\ast }(x)\to 0$ as $x\to \pm \mathrm{\infty }$. For $u,v\in H^1(\mathbb{R})$ the elementary bound $\mathrm{\mid }{u}^{\ast }\overline{v^{\ast }}\mathrm{\mid }\le \frac{1}{2}(\mathrm{\mid }{u}^{\ast }{\mathrm{\mid }}^2+\mathrm{\mid }{v}^{\ast }{\mathrm{\mid }}^2)$ and $u^{\ast }(x)\to 0$, $v^{\ast }(x)\to 0$ give $u^{\ast }(x)\overline{v^{\ast }(x)}\to 0$ as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$. This is (c).