# Show $W^{1,q}_0(-1,1)\subset C([-1,1])$

I need show that space $W^{1,q}_0(-1,1)$ is a subset of $C([-1,1])$ space. How I will able to doing this?

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If $q=+\infty$, and $\varphi_n$ are test functions such that $\lVert \varphi_n-u\rVert_{\infty}\to 0$, then we can find a set of measure $0$ such that $\sup_{x\in [-1,1]\setminus N}|\varphi_n(x)-u(x)|\to 0$, so $u$ is almost everywhere equal to a continuous function, and can be represented by it.
We assume $1\leqslant q<\infty$. As $W_0^{1,q}(-1,1)$ consists of equivalence classes of functions and $C[-1,1]$ consists of functions, what we have to show is that each element of $W_0^{1,q}(-1,1)$ can be represented by a continuous function. First, for $\varphi$ a test function, we have $$|\varphi(x)-\varphi(y)|\leqslant \left|\int_x^y|\varphi'(x)|dx\right|\leqslant |x-y|^{1-1/q}\lVert\varphi'\rVert_{L^q}.$$ Now, let $u\in W_0^{1,q}(-1,1)$. By definition, we can find a sequence $\{\varphi_k\}\subset D(-1,1)$ such that $\lim_{k\to+\infty}\lVert u-\varphi_k\rVert_q+\lVert u'-\varphi'_k\rVert_q=0.$ Up to a subsequence, we can assume that $\lim_{k\to+\infty}\varphi_k(x)=u(x)$. As for $k$ large enough, $$|\varphi_k(x)-\varphi_k(y)|\leqslant |x-y|^{1-1/q}(\lVert u'\rVert_{L^q}+1),$$ we have for almost every $x, y\in (-1,1)$, $$|u(x)-u(y)|\leqslant |x-y|^{1-1/q}(\lVert u'\rVert_{L^q}+1),$$ what we wanted (and even more, as $u$ is represented by a $\left(1-\frac 1q\right)$-Hölder continuous function, as noted robjohn).
So not only are they continuous, but they are also Hölder $1-1/q$ continuous. (+1) –  robjohn Nov 6 '12 at 17:00
@DavideGiraudo Why this $\lVert \varphi_n-u\rVert_{\infty}\to 0$. For any function $u$ is this true? –  user46060 Nov 7 '12 at 20:17
I meant, for $u\in W^{1,\infty}$, hence in particular in $L^{\infty}$. –  Davide Giraudo Nov 7 '12 at 20:19