In Luenbeger's book Optimization by Vector Space Methods, chapter 6, the adjoint of a linear operator is defined in the following way:

Let $X$ and $Y$ be normed spaces and let $A: X \mapsto Y$ be a bounded linear operator. The adjoint operator $A^*: Y^* \mapsto X^*$ is defined by the equation $\langle x, A^*y^* \rangle = \langle Ax, y^* \rangle$.

Where $X^*$ and $Y^*$ are the dual spaces of $X$ and $Y$, respectively. Then, the author proceeds saying (I have bolded the part I am concerned with):

This important definition requires a bit of explanation and justification. Given a fixed $y^* \in Y^*$, the quantity $\langle Ax, y^* \rangle$ is a scalar for each $x \in X$ and is therefore a functional on X. Furthermore, by the linearity of $y^*$ and $A$, it follows that this functional is linear. Finally, since \begin{equation} |\langle Ax, y^* \rangle| \leq ||y^*||\,||Ax^*|| \leq ||y^*||\,||A||\,||x^*|| \end{equation} it follows that this functional is bounded and is thus an element $x^*$ of $X^*$. We then define $A^*y^* = x^*$. The adjoint is obviously unique and the reader can verify that it is linear.

I am totally ok with linearity, but it is far from obvious to me that the adjoint (exists and) is unique.

I have searched other sources (e.g. https://web.eecs.umich.edu/~fessler/course/600/l/l06.pdf) and most of them further assume that $X$ and $Y$ are Hilbert spaces (i.e. there's some inner product defined there), and then existence and uniqueness can be proved using the Riesz representation theorem, a.k.a. Riesz-Fréchet theorem (still, I would not say that the result is "obvious").

How can we establish the result for general normed spaces? More importantly, why is this result so obvious?

  • $\begingroup$ But every normed linear space need not have an inner product! What does those innerproduct mean? $\endgroup$ – SunShine Nov 23 at 7:24
  • $\begingroup$ where do you see the inner product? $\endgroup$ – D... Nov 23 at 12:14
  • $\begingroup$ This Equation $\langle x, A^*y^* \rangle = \langle Ax, y^* \rangle$. $\endgroup$ – SunShine Nov 23 at 12:22
  • $\begingroup$ this is not an inner product. $\langle x, y^* \rangle$ is a notation for $y^*(x)$, i.e. for applying the functional $y^*$ to the vector $x$. $\endgroup$ – D... Nov 23 at 16:20

Here $\langle x,y^*\rangle$ stands for $y^*(x)$, so the adjoint is defined by $A^*y^*(x)=y^*(Ax)$. So we define $A^*$ as the mapping $y^*\mapsto(x\mapsto y^*(Ax))$. I hope this makes it more obvious that this is well-defined linear and unique.

  • $\begingroup$ got it! this notation helped a lot :) $\endgroup$ – D... Dec 14 '18 at 15:35

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