\[V\]
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 \[D\]
has no adjoint.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 \[D\]
has no adjoint.