$W_0^{1,1}([0,+\infty)) \subset L^\infty((0,\infty))$?

Question

If we define \begin{align*} W^{1,1}_0 (0, \infty) = \{ f \in W^{1,1}(0, \infty):\, f(0)=0\} \ , \end{align*} we can say that if $f \in W^{1,1}_0 (0, \infty) \implies f \in L^\infty (0,\infty)$? Do you know a counterexample? Thanks for your help

Answer 1

Suppose that $u \in C^\infty_0((0,\infty))$. For $0<x\leqslant y$, the Fundamental Theorem of Calculus implies that \begin{align*} \vert u(y) - u(x) \vert &= \bigg \vert \int_x^y u'(t) \, d t \bigg \vert \\ &\leqslant \int_x^y \vert u'(t) \vert \, d t \\ &\leqslant \| u' \|_{L^1((0,\infty))}. \end{align*} Sending $x \to 0^+$, implies that $$ \| u \|_{L^\infty((0,\infty))} \leqslant \| u' \|_{L^1((0,\infty))}. $$

By density of $C^\infty_0((0,\infty))$ in $W^{1,1}_0((0,\infty))$ the above inequality holds in $W^{1,1}_0((0,\infty))$.