Let $W_{2}^{1} = \lbrace f:[0,1] \to \mathbb{C} : f \in AC[0,1], f' \in L^{2}[0,1] \rbrace$ be a Hilbert space with an inner product: $\langle f, g \rangle = \int_{0}^{1} f\overline{g}dx + \int_{0}^{1} f'\overline{g'}dx$, and let $F: W_{2}^{1} \to \mathbb{C}$ be a linear functional defined by evaluation at zero: $F(f) = f(0)$.

I want to prove and obtain the following properties of $F$:

  1. I want to prove that $F$ is a bounded linear functional.
  2. I want to find the function $\phi$ s.t. $F(f) = \langle f, \phi \rangle$, and I want to find $||F||$.

I instantly hit a wall in this problem. After some googling, I found out that $W_{2}^{1}$ is a kind of Sobolev space; however, this is a problem from functional analysis course, and Sobolev spaces haven't been covered. I think that this problem has a solution which doesn't use any properties of Sobolev spaces.

I was able to prove that $F$ is bounded, albeit after some trouble:

\begin{equation} |F(f)| = |f(0)| = |f(t) - \int_{0}^{t} f'(s)ds| \leq |f(t)| + \int_{0}^{t} |f'(s)|ds \leq |f(t)| + \int_{0}^{1} |f'(s)|ds \hspace{0.5cm}(1)\end{equation}

Now, since $f$ is continuous, there exists a $t_{0}$ such that $m = |f(t_{0})| = \min_{t \in [0,1]}|f(t)|$. Then $m \leq \int_{0}^{1} |f(t)|dt$. Since $(1)$ holds for all $t \in [0,1]$, by inserting $t=t_{0}$ into $(1)$, we get

\begin{equation} |f(0)| \leq m + \int_{0}^{1}|f'(s)|ds \leq \int_{0}^{1} |f(s)|ds + \int_{0}^{1} |f'(s)|ds \leq \sqrt{\int_{0}^{1} |f'(s)|^2ds} + \sqrt{\int_{0}^{1} |f(s)|^2ds} \leq \sqrt{2}||f||. \end{equation}

For the next-to-last inequality I used Cauchy-Schwartz, and for the last inequality I used $\sqrt{a} + \sqrt{b} \leq \sqrt{2}\sqrt{a+b}$, which follows from the AM-GM inequality.

Therefore, $||F|| \leq \sqrt{2}$.

However, I'm not sure how to find $\phi$ with $F(f) = f(0) = \langle f, \phi \rangle$. If I were to find such a $\phi$, then obviously by the Riesz representation theorem, $||F|| = ||\phi||$. I took a look at Find $y \in W_{2}^{1}[-1,1]$ s.t. $\forall x \in W_{2}^{1}[-1,1]$, $f(x)=\langle x, y \rangle$, but the operator there is very different from mine, and I can't get a differential equation out of the condition analogous to the one arctic tern obtained there. Is there a good "algorithm" for finding the corresponding vectors to Hilbert space functionals? Or at least in Hilbert spaces with integral inner products?


What you want is a function $\varphi$ so that

$$\tag{1} \int_0^1 f\varphi + \int_0^1 f' \varphi' = f(0), \ \ \ \forall f\in W^2_1.$$

The left hand side is (assuming that $\varphi$ is twice differentiable)

$$ \int_0^1 f\varphi - \int_0^1 f\varphi'' + f(1) \varphi' (1) - f(0) \varphi '(0).$$

Thus if $\varphi$ satisfies

\begin{cases} \varphi'' = \varphi, \\ \varphi'(0) = -1, \ \ \varphi'(1) = 0. \end{cases}

Then one has (1). This is a second order differential equation has one can solve easily.

  • 1
    $\begingroup$ Oh, right! The answer from the linked question actually works... I just wasn't careful enough with choosing the boundary conditions. $\endgroup$ – Matija Sreckovic Jun 15 '18 at 17:26
  • 1
    $\begingroup$ I think that it's $f(1)\varphi'(1) - f(0)\varphi'(0)$ in the second equation, giving boundary conditions $\varphi'(0) = -1, \varphi(1) = 0$. My final solution is $\varphi = \frac{1}{e^2-1}e^x + \frac{e^2}{e^2-1}e^{-x}$, after solving the DE. $\endgroup$ – Matija Sreckovic Jun 15 '18 at 17:31
  • $\begingroup$ Yes indeed. Corrected. $\endgroup$ – user99914 Jun 15 '18 at 17:32

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