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  
\[D\]
  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.

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