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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.

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