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I am trying to prove that if $u:(0,1)\to\mathbb{R}$ lies in $W^{1,1}(0,1)$, then $u\in C(0,1)$. Is there any help anybody can offer?

Thanks.

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    $\begingroup$ I don't think that your question fits the scope of MO. MSE (math.stackexchange.com) would have been a better choice. $\endgroup$
    – Salvo Tringali
    Commented Mar 9, 2013 at 16:33
  • $\begingroup$ Apologies, I didn't realise the difference. $\endgroup$
    – John
    Commented Mar 9, 2013 at 16:46

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Since $u'\in L^1(0,1)$, you find from the Lebesgue differentiation theorem that $$ \int_{1/2}^x u'(t) dt=u(x)+Cst,\quad x\in(0,1). $$ As a result $u$ is a continuous function and the constant above is $-u(1/2).$

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  • $\begingroup$ that's it. it could be additionally mentioned that $W^{1,1}$-functions are also absolutely continuous. $\endgroup$
    – Delio Mugnolo
    Commented Mar 9, 2013 at 16:29

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