# Existence of adjoint in normed space

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?

• But every normed linear space need not have an inner product! What does those innerproduct mean? – SunShine Nov 23 at 7:24
• where do you see the inner product? – D... Nov 23 at 12:14
• This Equation $\langle x, A^*y^* \rangle = \langle Ax, y^* \rangle$. – SunShine Nov 23 at 12:22
• 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 $x$. – 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.