Does $H^1(0,\infty)$ function posses a uniformly continuous representative?

Question

I need to prove that a function from the Sobolev space $H^1(0,\infty)$ tends to zero as $x$ tends to infinity. I was able to prove that if a function is uniformly continuous, the limit must be zero, so it would be enough to prove that a function from $H^1(0,\infty)$ posseses a uniformly continuous representative. I know that for $-\infty < a < b < \infty$, function from $H^1(a,b)$ has an absolutely continuous (therefore uniformly continuous) representative, can this be generalized to my case where interval is infinite?

Answer 1

Consider $0<x<y$, and use the fundamental theorem of calculus to get $$ |f(y)-f(x)|= \left| \int_x^y f'(z)\, dz \right| \leq |y-x|^{1/2} \left(\int_{x}^y |f'(z)|^2\, dz \right)^{1/2}, $$ where we used Hölder's inequality for the last step. This implies $f$ is Hölder continuous with exponent $1/2$, and in particular it's uniformly continuous. Notice that the above inequality is valid for $x,y$ in any interval on the real line, bounded or not.