## Adjoint of Inner Product Example

Let
$V$
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
$V$
$T^*$
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{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}