Find element in Riesz Representation Theorem in a Sobolev Space

Question

I am trying to solve the following question:

Let $H = H^1([0,1])$ and let $Tu = u(1)$. Let $\langle \cdot,\cdot \rangle_H$ denote the standard inner product in $H$. By the Riesz representation theorem, there exists a unique $v\in H$ such that $Tu = \langle u, v \rangle_H$ for all $u\in H$. Find $v$.

Here is my partial proposed solution:

Note that since $H^1( [0,1] ) \subset C^{0, 1/2} ( [0,1] ) $, for each $u \in H^1( [0,1] )$ we can consider a periodic extension to all of $\mathbb{R}$. Using the Fourier series decomposition of a $1$-periodic function, $H^1([0,1])$ is isometrically isomorphic to the periodic Sobolev space H_{1}^1([0,1]) consisting of $f \in L^2([0,1])$ such that

$$||f||^2_{H_{1}^1}=\sum_{n \in \mathbb{Z}}( 1 + (2 \pi n)^2 )| \widehat{f_n} |^2 < \infty$$

Here $( \widehat{f_n} )$ denotes the sequence of Fourier coefficients of $f$. $H^1_1$ can be endowed with the inner product:

$$\langle f, g\rangle_{H^1_1}=\sum_{n \in \mathbb{Z}}( 1 + (2 \pi n)^2 )\widehat{f_n} \widehat{g_n} $$

We find $v \in H_{1}^1$ such that:

$$ u(1)=\langle u, v \rangle_{H_{1}^1}=\sum_{n \in \mathbb{Z}}( 1 + (2 \pi n)^2 )\widehat{u_n} \widehat{v_n} $$

Let $v \in H^1([0,1])$ such that $\widehat{v}_n = \frac{1}{ 1 + (2 \pi n)^2} $, then:

$$ \langle u, v \rangle_{H_{1}^1}=\sum_{n \in \mathbb{Z}}\widehat{u_n} =\sum_{n \in \mathbb{Z}}e^{2 \pi i n}\widehat{u_n} =u(1). $$

Hence we can take:

$$ v(x)=\sum_{n \in \mathbb{Z}}\frac{e^{2 \pi i x n}}{ 1 + (2 \pi n)^2} $$

Suggestions? I don't know how to further simplify $v(x)$.

Answer 1

The equation defining the Riesz-representative is $$ \int_0^1 u'v' + uv \ dx = u(1) \quad \forall u\in H^1. $$ This is the weak formulation of the boundary value problem $$ -v'' + v = 0 \text{ in } (0,1) $$ with boundary conditions $v'(0)=0$ and $v'(1) = 1$. Its solution is given by wolframalpha as $$ y(x) = \frac{e^{1 - x} + e^{x + 1}}{e^2 - 1} = \frac{ \cosh x}{\sinh 1}. $$