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I would like to know whether the the Fundamental theorem of calculus (Part II) can be applied in the following setting.

Let $(a,b)$ be an open interval in $R^1$.

Let $u \in H^1((a,b))$ with $u(a)=0$

Then, I know by Sobolev Embedding Theorem that $u$ is in fact in $C([a,b])$

Now, what I am wondering is whether I can write

$u(b)= u(a)+\int^b_au'(x)dx=\int^b_au'(x)dx$

It seems true to me, but how do I actually prove it?

mononono
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    Strictly speaking, $u(a)$ and $u(b)$ are not defined yet. You should work with $u(x)$ where $x\in (a,b)$ is close to an endpoint. –  Feb 13 '15 at 02:05

1 Answers1

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I'll answer in slightly greater generality, without using $u(a)=0$.

Since you already know the (continuous) embedding $H^1\to C((a,b))$, argue by density. Let $u_n$ be a sequence of smooth functions converging to $u$ in $H^1$. Then $u_n\to u$ also in $C((a,b))$, i.e., uniformly (if we use the continuous representative of $u$). Also, $u_n'\to u'$ in $L^2$, hence in $L^1$, by Hölder's inequality.

So, for any points $s,t$ with $a<s<t<b$ we have $$ u(t)-u(s) = \lim_{n\to\infty}u_n(t)-u_n(s) = \lim_{n\to\infty}\int_t^s u_n'(\xi)\,d\xi = \int_t^s u'(\xi)\,d\xi $$ Let $s,t$ approach endpoints (one at a time) and conclude that $u$ has finite limits at both endpoints.