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## One dimensional trace of a Sobolev function

Check that the mapping ${u\mapsto u(0)}$ from ${H^1(\Bbb{R})}$ to ${\Bbb{R}}$ is a continuous linear functional on ${H^1(0,1)}$. Deduce that there exists a unique ${v_0 \in H^1(0,1)}$ such that

$\displaystyle u(0)=\int_0^1(u'v_0'+uv_0).$

Show that ${v_0}$ is the solution of some differential equation with appropriate boundary conditions and compute ${v_0}$ explicitly.

H. Brezis, Functional Analysis, Ex 8.18

Solution: Theorem 8.2 from the same book states that for ${u \in W^{1,p}(I)}$ with ${1\leq p\leq \infty}$ (${I}$ is an interval) then there exists a unique function ${\tilde u \in C(\overline{I})}$ such that ${u=\tilde u}$ a.e. on ${I}$ and

$\displaystyle \tilde u(x)-\tilde u(y)=\int_y^x u'(t)dt,\ \forall x,y \in \overline{I}.$

The proof is presented in the book, but the main idea is that every function ${u \in W^{1,p}(I)}$ has a continuous representative and we have a nice integral representation for ${u(x)-u(y)}$. (Note that this does not happen in higher dimensions) In the following I will use always the continuous representative of a function in ${H^1(0,1)=W^{1,2}(0,1)}$. Suppose we have ${u,v \in H^1(I)}$ with ${I=(0,1)}$. Using the above formula we get

$\displaystyle u(x)=u(0)+\int_0^x u'(t)dt$

$\displaystyle v(x)=v(0)+\int_0^x v'(t)dt$

and therefore we have

$\displaystyle |u(0)-v(0)|\leq |v(x)-u(x)|+\left|\int_0^x [u'(t)-v'(t)]dt\right|.$

Using Holder’s inequality we get

$\displaystyle |u(0)-v(0)|\leq |v(x)-u(x)|+\|u'-v'\|_{L^2(I)}$

Integrating the last inequality on ${I}$ and using again Holder’s inequality we obtain

$\displaystyle |u(0)-v(0)|\leq \|u-v\|_{L^2(I)}+\|u'-v'\|_{L^2(I)}.$

This last inequality proves that the considered application is continuous. The linearity is trivial to prove, therefore ${u \mapsto u(0)}$ is a continuous functional. Because ${H^1(0,1)}$ is a Hilbert space we can apply Riesz representation theorem to conclude that there exists a unique ${v_0 \in H^1(I)}$ such that

$\displaystyle u(0)=\langle u,v_0\rangle_{H^1(I)}=\int_0^1(u'v_0'+uv_0), \forall u \in H^1(I).$

It’s not hard to see that ${v_0}$ is the solution of

$\displaystyle \begin{cases} -v_0''+v_0=0 \\ v_0'(0)=-1 \\ v_0'(1)=0 \end{cases}$

which is not hard to solve.