\[V\]
be the vector space of polynomials over the field of complex numbers with inner product \[\]
\[f \cdot g= \int^1_0 f(t) \bar{g} (t) dt\]
(where \[\bar{g} (t)\]
) is the complex conjugate of \[g(t)\]
. What is the adjoint of the linear operator
\[M_f(g)=fg\]
(multiplication by \[f\]
?)If
\[V\]
is a linear product space and \[T\]
a linear operator on \[V\]
the adjoint \[T^*\]
of \[T\]
is a linear operator such that \[< T \alpha , \beta > = < \alpha , \beta T^* >\]
for all \[\alpha , \; \beta \in V\]
.We can write
\[f(x)= \sum_k a_k x^k , \; \bar{f}(x)= \sum_k \bar{a}_k x^k\]
.If
\[f(x)= \bar{f}(x)\]
then \[f\]
is real.The adjoint of 'multiplication by
\[f\]
' is 'multiplication by \[\bar{f}\]
'.\[\begin{equation} \begin{aligned} M_f (g) \cdot h &= \int^1_0 (fg)(t) \bar{h}(t)dt \\ &= \int^1_0 f(t)g(t) \bar{h}(t)dt \\ &= \int^1_0 g(t) \bar{ \bar{f} h}(t)dt \\ &= < g , \bar{f}h > \\ &= g \cdot M_{\bar{f} h} \end{aligned} \end{equation}\]