= Solution
For the <Volterra integration operator>, <Cauchy-Schwarz inequality> gives
$$
|Jf(x)|^2=\left|\int_0^x f(t)\,dt\right|^2
\leq x\int_0^x|f(t)|^2\,dt.
$$
Integrating and reversing the order yields
$$
\|Jf\|_2^2\leq\int_0^1\frac{1-t^2}{2}|f(t)|^2\,dt
\leq\frac12\|f\|_2^2.
$$
Thus $J$ is bounded, with $\|J\|\leq1/\sqrt2$. Since $f\in L^2(0,1)\subseteq L^1(0,1)$, $Jf$ is absolutely continuous and $(Jf)'=f$ almost everywhere. If $Jf=0$ as an $L^2$ element, its continuous representative is zero everywhere, so its <derivative> is zero almost everywhere. Hence \b[$J$ is one-to-one].
Its range is exactly $\{u\in H^1(0,1):u(0)=0\}$, and $Au=iu'$. This contains the smooth compactly supported functions in $(0,1)$, so $A$ is a <densely defined operator>. Closedness can also be seen directly: if $u_n\to u$ and $Au_n\to v$ in $L^2$, then $u_n=J(-iAu_n)\to J(-iv)$, giving $u=J(-iv)\in D(A)$ and $Au=v$.
For every $\lambda\in\mathbb C$,
$$
(\lambda I-A)J=\lambda J-iI.
$$
If $\lambda\ne0$, the allowed fact $\sigma(J)=\{0\}$ makes $J-(i/\lambda)I$ invertible. At $\lambda=0$, $\lambda J-iI=-iI$ is invertible too. Thus the <bounded operator>
$$
R_\lambda=J(\lambda J-iI)^{-1}
$$
has range in $D(A)$ and is a two-sided inverse of $\lambda I-A$. Equivalently, solving the initial-value differential equation gives
$$
R_\lambda h(x)=i\int_0^x e^{-i\lambda(x-s)}h(s)\,ds.
$$
This is bounded for every fixed $\lambda$ on the finite interval. Therefore <Volterra inverse differentiation has empty spectrum>:
$$
\boxed{\sigma(A)=\varnothing.}
$$
The familiar nonempty-spectrum theorem for bounded operators does not apply to this unbounded operator.
The subspace $H_0$ is the kernel of the bounded functional $f\mapsto Jf(1)=\int_0^1f$, so it is closed. Its image under $J$ is
$$
D(A_0)=\{u\in H^1(0,1):u(0)=u(1)=0\}=H_0^1(0,1),
$$
the <zero-trace Sobolev space>. If $u_n\in D(A_0)$, $u_n\to u$ and $A_0u_n\to v$, then $-iA_0u_n\in H_0$ converges to $-iv\in H_0$, and $u=J(-iv)\in D(A_0)$. Thus $A_0$ is closed. It is densely defined because it also contains the smooth compactly supported functions.
For $u,v\in D(A_0)$, <integration by parts> and their zero endpoint traces give
$$
\langle A_0u,v\rangle
=i\int_0^1u'\overline v
=i[u\overline v]_0^1-i\int_0^1u\overline{v'}
=\langle u,A_0v\rangle.
$$
Hence \b[$A_0$ is closed and symmetric].
For any $v\in C^1([0,1])$, the same formula gives $\langle A_0u,v\rangle=\langle u,iv'\rangle$, with no endpoint term because $u$ vanishes there. Consequently $v\in D(A_0^*)$ and $A_0^*v=iv'$, proving the requested inclusion. In fact, testing against compactly supported smooth $u$ shows that every element of $D(A_0^*)$ has <weak derivative> in $L^2$; conversely the integration-by-parts formula applies to every $v\in H^1(0,1)$. Thus
$$
D(A_0^*)=H^1(0,1),\qquad A_0^*v=iv',
$$
with no boundary restrictions.
For any $\lambda\in\mathbb C$, the nonzero function $v_\lambda(x)=e^{-i\lambda x}$ belongs to this domain and satisfies $iv_\lambda'=\lambda v_\lambda$. Conversely this first-order equation has only scalar multiples of that exponential. Therefore
$$
\boxed{\sigma_p(A_0^*)=\mathbb C,\qquad
\ker(A_0^*-\lambda I)=\operatorname{span}\{e^{-i\lambda x}\}.}
$$
For any $\lambda$, the adjoint eigenfunction with <eigenvalue> $\overline\lambda$ annihilates $\operatorname{Ran}(\lambda I-A_0)$, so this range is not all of $H$. Explicitly the resolvent equation with zero initial value has the solution $R_\lambda h$ above, and its endpoint value is zero exactly when
$$
\int_0^1e^{i\lambda s}h(s)\,ds=0.
$$
Thus the range is a proper closed hyperplane for every $\lambda$. This proves that the <two-endpoint symmetric derivative has whole complex spectrum>:
$$
\boxed{\sigma(A_0)=\mathbb C.}
$$
There are no <eigenvalues> of $A_0$ itself: the eigenfunction equation together with $u(0)=0$ forces the exponential coefficient to vanish. The operator is symmetric but not self-adjoint, since its adjoint has a strictly larger domain; symmetry alone does not force a real <spectrum>.
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