Sobolev Injections

Question

Is it true that $W^{1,\infty}(]0,\infty[) \hookrightarrow C([0,\infty[)$ and $W^{1,1}(]0,\infty[) \hookrightarrow C([0,\infty[)$ ? Thanks

Answer 1

So the first one is very simple. $$ f(x)-f(y) = \int_x^y Df(t) \text{d} t \leq \| Df \|_{L^\infty(0,\infty)} \left| \int_x^y \text{d} t \right| \leq \| Df \|_{L^\infty(0,\infty)}\left|x- y \right|, $$ so $f$ is not only continuous, it is Lipschitz.

For the second one, you need to work a little more, but the idea is similar. For any $x>0$ because $Df$ is integrable $$ \lim_{n\to\infty} \int_{x-\frac 1n}^{x+\frac 1n} \left| Df((t) \right| \text{d}t =0. $$ Thus for any $\epsilon>0$, there exists $N$ such that for all $n\geq N$ $$ \int_{x-\frac 1n}^{x+\frac 1n} \left| Df((t) \right| < \epsilon. $$ Thus, for any $y\in(x-\frac 1N,x+\frac 1N)$ there holds $$ \left| f(x)-f(y) \right| \leq \left| \int_x^y \left| Df(t) \right|\text{d} t \right|\leq \int_{x-\frac 1n}^{x+\frac 1n} \left| Df((t) \right| < \epsilon. $$