From my understanding, any inner product on $\mathbb{F}^n$ may be defined in the following form (citation):

$$\langle x,y \rangle = y^t M x$$ where $M$ is a positive definite, hermitian matrix.

We can then determine the adjoint of a linear operator $T$, with respect to the above inner product $T^*:=M^{-1}A^*M$

Now say we are working in the space of complex-valued functions $V$, where $T:V\to V$ is defined by $T(f) = hf$ $$\langle f,g\rangle = \int_{0}^{1} f(t)\overline{g(t)}dt$$

where $h \in V$.

How would we determine the adjoint of such an operator? Is there a similar method to the one mentioned above?

  • $\begingroup$ What is $A^*$ in the definition of $T^*$? $\endgroup$
    – stochastic
    Dec 10, 2017 at 4:25
  • $\begingroup$ $A$ would be the matrix representation of $T$, and $A^*$ would be the conjugate transpose of $A$. $\endgroup$
    – Taylor Tam
    Dec 10, 2017 at 4:33

2 Answers 2


If I understand correctly, you are trying to define $T^*$ so that $$\langle Tx,y\rangle=\langle x,T^*y\rangle.$$

If that is the case, $$\langle T(f),g\rangle=\langle hf,g\rangle=\int_0^1 f(t)h(t)\overline{g(t)}dt =\int_0^1 f(t)\overline{\overline{h(t)}g(t)}dt=\langle f,\overline hg\rangle\implies T^*(g) = \overline hg.$$


Assume that $T:L^{2}\rightarrow L^{2}$. Let $\varphi:L^{2}\rightarrow(L^{2})^{\ast}$ be the canonical map, that is, $\left<f,\varphi(g)\right>=\displaystyle\int f\overline{g}$, then consider the Banach adjoint $T^{\ast}:(L^{2})^{\ast}\rightarrow(L^{2})^{\ast}$ and let $T'=\varphi^{-1}\circ T^{\ast}\circ\varphi$, if it were $\overline{h}g\in L^{2}$, then \begin{align*} \left<f,\varphi\left(\overline{h}g\right)\right>&=\int f\overline{\overline{h}g}\\ &=\int fh\overline{g}\\ &=\int T(f)\overline{g}\\ &=\left<T(f),\varphi(g)\right>\\ &=\left<f,(T^{\ast}\circ\varphi)(g)\right>\\ &=\left<f,(\varphi\circ T')(g)\right>, \end{align*} then $(\varphi\circ T')(g)=\varphi\left(\overline{h}g\right)$, so $T'(g)=\overline{h}g$. For example, if $h$ is a Schwartz function, then it goes through.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .