## An Operator Without an Adjoint

Let
$V$
be the vector space of polynomials with complex coefficients with the inner product
$\lt f(t), \; g(t) \gt =^1_0 f(t)g(t)dt$

Let
$D$
be the differential operator. Then
$D$
Using integration by parts
$\int^1_0 (Df)gdt=f(1)g(1)-f(0)g(0)-f(Dg)$

Let
$g$
be a fixed function. Assume that
$D$
has an adjoint, so there is a polynomial
$D^*g$
such that
$\lt Df, \; g \gt = \lt f, D*g \gt$
for all
$f \in V$
.
Then
$\lt f, \; D^*g \gt = f(1)g(1)-f(0)g(0)$

With
$g$
fixed,
$L(f)=f(1)g(1)-f(0)g(0)$
is a linear functional and cannot be of the form
$L(f)= \lt f, \; h \gt$
unless
$L=0$
. If
$D^*g$
exists, then with
$h=D^*g+Dg$
we do have
$L(f)= \lt f, \; h \gt$
and so
$g(0)=g(1)=0$
.
The existence of a suitable polynomial
$D^*g$
implies that
$g(0)=g(1)=0$
.
Conversely, if
$g(0)=g(1)=0$
then
$D^*g=-Dg$
satisfies
$\lt Df, \; g \gt = \lt f, \; D^*g \gt$
for
$f \in V$
. If
$g(0), g(1) \neq 0$
then
$D^*g$
cannot be suitably defined, so
$D$