Question concerning proof of unique existence of adjoint operator

Question

Here, a proof is given on p. 50. However, one crucial moment is when they define $$T^{\star}k := y.$$ Okay, the next step reads: $$\langle h, T^{\star}\left( \lambda_1k_1 + \lambda_2k_2 \right)\rangle_{H} = \langle Th, \lambda_1 k_1 + \lambda_2 k_2 \rangle_{K},$$ which I don't understand. Could somebleady please explain this step in more detail? I know that it previously stated $$\langle Th, k\rangle_{K} = \langle h, y \rangle_{H} \ \forall h \in H,$$ but I still don't see it yet.

Answer 1

Here $k=\lambda_1k_1+\lambda_2k_2$, $y=T^*k$.

Answer 2

The Theorem 6.18 at page 50 introduces the reader to the adjoint operator $T^*:K \to H$ of an operator $T: H \to K$. In short words, its property is:

$$\left\langle Th, k\right\rangle_K = \left\langle h, T^*k\right\rangle_H.$$

By the Riesz-Frechet representation theorem, there exists $y \in H$, such that:

$$\left\langle Th,k\right\rangle_K = \left\langle h,y\right\rangle_H.$$

Since $T^*: K \to H$, thus $T^*(\cdot) \in H$. Assume then, that $ H\ni y = T^*k \in H$. Then, you get the expression:

$$\left\langle Th, h\right\rangle_K = \left\langle h,T^*k\right\rangle_H.$$

Then, let $k= \lambda_1 k_1 + \lambda_2 k_2$ and thus $T^*k = T^*(\lambda_1k_1 + \lambda_2 k_2) = y$. Proceed with using the adjoint property, as:

\begin{align} \left\langle h,y\right\rangle =\langle h, T^* (\lambda_1 k_1 + \lambda_2 k_2) \rangle &= \langle h, T^\ast k \rangle \\ &= \langle Th, k \rangle \\ &= \langle Th, \lambda_1 k_1 + \lambda_2 k_2 \rangle. \end{align}