It is true to write that $W^{1,\infty}(]0,\infty[) \hookrightarrow C([0,\infty[)$ et $W^{1,1}(]0,\infty[) \hookrightarrow C([0,\infty[)$ ? Thanks
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$\begingroup$ What is the question? $\endgroup$– JRNCommented Apr 5, 2021 at 13:36
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2$\begingroup$ @Bazin, your edit significantly alters the original question: in your edit the interval becomes relatively compact (this is significant and much easier than the original version!), and you drop $W^{1, \infty}$ altogether. I must fight the temptation of rolling it back. $\endgroup$– Alex M.Commented Apr 5, 2021 at 13:52
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$\begingroup$ @JoelReyesNoche: Always look at the edit history of a post when it seems not to make sense anymore. $\endgroup$– Alex M.Commented Apr 5, 2021 at 13:53
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$\begingroup$ @AlexM. Thanks. I suggest you roll back the edit. $\endgroup$– JRNCommented Apr 5, 2021 at 13:56
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2$\begingroup$ @AlexM. I on the other hand succumbs to said temptation. $\endgroup$– Willie WongCommented Apr 5, 2021 at 14:26
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Let $f$ be a fonction in $L^1(0,+\infty)$ with norm 1 and let us define for $x\ge 0$ $$ \phi(x)=\int_x^1f(t) dt. $$ Then $\phi$ is continuous since $ -\phi(h)+\phi(0)=\int_0^h f(t) dt, $ which goes to 0 with $h$ from the Lebesgue Dominated Convergence Theorem.
