Determining the domain of the adjoint of $T = i\frac{d}{dx}$ on $C^1[0,1] \subseteq L^2[0,1]$

Question

I have read this post Distinguishing between symmetric, Hermitian and self-adjoint operators

But I have some specific questions relating to the operator $$T : \psi \longmapsto i \frac{d}{dx} \psi(x).$$ This operator is symmetric, meaning that $\langle \psi, T\varphi \rangle = \langle T\psi, \varphi \rangle$ for all $\varphi, \psi \in D(T)$. I'm aware that $T$ will be self-adjoint if $D(T^{\ast}) \subset D(T)$. My question is, since we know that $T$ is symmetric, does that immediately give us that the formula for $T^{\ast}$ will be $T$ itself?

Moreover, if this is true, how do we determine the domain of $T^{\ast}$. For example, suppose the domain of $T$ is given to be $\psi \in C^1[0,1]$ with vanishing boundary conditions $\psi(0) = \psi(1) =0$. How do we determine the domain $D(T^{\ast}$)?

Answer 1

My question is, since we know that $T$ is symmetric, does that immediately give us that the formula for $T^*$ will be $T$ itself?

That $T$ is symmetric means not only that $D(T) \subseteq D(T^*)$ but also that $T^* x = Tx$ on $D(T)$, i.e. $T^*$ is an extension of $T$. In fact, this is a characterization:

Proposition. Let $T$ be densely defined with domain of definition $D$. Then the following are equivalent:

1. $\langle T x, y \rangle = \langle x, Ty \rangle$ for all $x,y \in D$.

2. $T \subseteq T^*$, i.e. $D(T) \subseteq D(T^*)$ and $T y = T^* y$ for all $y \in D$.

If one of the previous conditions is fulfilled one calls $T$ symmetric.

Proof. (1.) $\implies$ (2.): Let $y \in D(T)$. Then the linear functional $\varphi_T(x) := \langle Tx, y\rangle$ is continuous by Riesz-Frechet, since $\varphi_T(x) = \langle x, Ty \rangle$ by (1.). This gives $y \in D(T^*)$. But then, for all $x \in D$ we have $$ \langle x, Ty \rangle = \langle Tx, y \rangle = \langle x, T^* y \rangle. $$ Since $D$ is dense, this gives $Ty = T^*y$ and the claim (2.) holds. The other implication is trivial. $\square$

This proof shows that $T^* = i \frac{d}{dx}$ on the domain of $D(T^*)$. Unless the operator $T$ is self-adjoint the domain of $T^*$ will be stricly greater than $D(T)$. Furthermore, there is already a definition of $D(T^*)$, namely $$ D(T^*) := \{ y \in H \colon x \mapsto \langle T x, y \rangle \text{ is continuous }\}. $$

I don't think there is a general rule for explicitely determining the domain of the adjoint. However in the case of the operator $T$ with domain $$D(T) := \{ f \in C^1[0,1] \colon f(0) = f(1) = 0\} \subseteq L^2[0,1]$$ we are lucky.

Claim. $D(T^*) = \mathrm{AC}[0,1]$ and $T^* = i \frac{d}{dx}$ on $D(T^*)$ (Note that the second statement is not automatic). Here $\mathrm{AC}$ denotes the absolutely continuous functions.

Proof. Let $f \in D(T^*)$ and $g := T^* f$. Consider the function $$ F(t) := \int_0^t g(s) ds. $$ Since $g \in L^2[0,1] \subseteq L^1[0,1]$, where the inclusion comes from the fact that we are dealing with a finite measure space, the function $F$ is absolutely continuous and hence a.e. differentiable with $ F' = g$ a.e. Hence, for all $h \in D(T)$ we have $$ \langle Th, f \rangle = \langle h, T^* f \rangle = \langle h, g\rangle = \langle h, F' \rangle = -\langle h', F \rangle \\ = -\frac{i}{i}\langle h', F \rangle = i \langle Th, F\rangle = \langle Th , -iF\rangle. $$ This gives us $$ \langle Th, f + iF \rangle = 0, $$ which leads to $f + iF \in (\operatorname{range} T)^\perp.$ By the fundamental theorem of calculus and the definition of $D(T)$, we know that $$ \operatorname{range} T = \left\{ f \in \mathrm{C}[0,1] \colon \int_0^1 f(t) = 0 \right\} = \left\{ f \in \mathrm{C}[0,1] \colon \langle f, 1\rangle = 0 \right\}. $$ This yields $$ \overline{\operatorname{range} T} = \{1\}^\perp $$ which in turn gives us $$ \{\operatorname{range} T\}^\perp = \overline{\operatorname{range} T}^\perp = \{1\}^{\perp\perp} = \operatorname{span} \{1\}. $$

Since $F$ and $1$ are absolutely continuous this gives us that $f = -iF + \alpha \cdot 1$ is absolutely continuous, i.e. $f \in \mathrm{AC}[0,1]$ and $$ i \frac{df}{dx} = F' = g = T^* f. $$ Now we show $D(T^*) \supseteq \mathrm{AC}[0,1]$. Let $f \in \mathrm{AC}[0,1]$ and $f' \in L^2$ be the a.e. existing derivative. Then, like above we calculate for all $g \in D(T)$ $$ \langle Tg, f\rangle = i\langle g',f\rangle = -i\langle g, f'\rangle = \langle g, if' \rangle,i.e. $$ the linear form $g \mapsto \langle Tg, f\rangle$ is continuous and hence $g \in D(T^*)$. $\square$

Answer 2

Maybe this argument works?

Integrating by parts, we see that $$T^{\ast} \psi = i \frac{d}{dx}\psi.$$ Now let $j(x)$ be any positive, smooth function with suppose $(-1,1)$ such that $\int_{-\infty}^{\infty} j(x) =1$. Define $j_{\epsilon}(x) := \epsilon^{-1} j(x/\epsilon)$. Fix $0 < \alpha < \beta < 1$ and define \begin{eqnarray*} f_{\epsilon}^{\alpha, \beta}(x) &=& j_{\epsilon}(x - \beta) - j_{\epsilon}(x- \alpha), \\ g_{\epsilon}^{\alpha, \beta}(x) &=& \int_0^x f_{\epsilon}^{\alpha, \beta}(t)dt. \end{eqnarray*} Let $\psi \in D(T^{\ast})$. For $\epsilon$ sufficiently small, $g_{\epsilon}^{\alpha, \beta} \in D(T)$ and $$(Tg_{\epsilon}^{\alpha, \beta}, \psi ) = (g_{\epsilon}^{\alpha, \beta}, T^{\ast}\psi ).$$ As $\epsilon \to 0$, $g_{\epsilon}^{\alpha, \beta} \to \chi_{(\alpha, \beta)} \in L^2(0,1)$, so $$(g_{\epsilon}^{\alpha, \beta}, T^{\ast} \psi) \longrightarrow - \int_{\alpha}^{\beta} (T^{\ast}\psi)(x) dx.$$ Moreover, if $\varphi \in C[0,1]$, then $$J_{\epsilon}\varphi = \int_0^1 j_{\epsilon}(x-t) \varphi(t)dt \xrightarrow{ \ \ L^2 \ \ } \varphi.$$ Further, we claim that $\| J_{\epsilon} \| \leq 1$. To see this, suppose that $\psi \in L^2[0,1]$, then \begin{eqnarray*} \left| (\psi, J_{\epsilon}\varphi) \right| & \leq & \iint j_{\epsilon}(x-t) \left| \varphi (t) \right| \left| \psi (x) \right| dx dt \\ &=& \iint j_{\epsilon}(y) \left| \varphi (t) \right| \left| \psi(y+t) \right| dy dt \\ & \leq & \| \varphi \| \| \psi \| \int j_{\epsilon}(y) dy \\ &=& \| \varphi \| \| \psi \|. \end{eqnarray*} By a simple $\epsilon/3$ argument, $J_{\epsilon} \varphi \to \varphi$ in $L^2$ for all $\varphi \in L^2[0,1]$. So $(Tg_{\epsilon}^{\alpha, \beta}, \psi) \to -i(\psi(\beta)- \psi(\alpha))$ in mean square as $\epsilon \to 0$. Therefore, for almost every $\alpha, \beta$, $$i(\psi(\beta) - \psi(\alpha)) = \int_{\alpha}^{\beta} (T^{\ast}\psi)(x)dx.$$ So $\psi$ is absolutely continuous and $$i \frac{d}{dx}\psi(x)= (T^{\ast}\psi)(x).$$ Conversely, integrating by parts, we see that if $\psi \in AC[0,1]$, then $\psi \in D(T^{\ast})$ and $T^{\ast} \psi = i \frac{d}{dx}$. We therefore conclude that $T^{\ast} = i d/dx$ and $D(T^{\ast}) = AC[0,1]$.

Reference: R&S 1.