I'm trying to prove the existence of the linear map $f^*$ in the following theorem about the adjoint:
Let $f: V \to W$ be a linear map, with $V$ finite-dimensional and $V,W$ inner product spaces over the same field. Then there is a unique linear map $f^*: W \to V$ such that $$\langle f(v),w \rangle = \langle v, f^*(w)\rangle$$ for all $v\in V, w \in W$.
Here's how far I got so far.
Try 1:
Fix $w \in W$. Because $V$ is finite-dimensional there exists a unique $x \in V$ for all linear forms $\psi_1 = \langle f(\cdot),w \rangle \in V^*$ — namely we have the isomorphism $V \to V^*$. Since the linear form $\langle f(\cdot),w \rangle$ is uniquely determined by $w \in W$, define the linear map $f^*$ by $x = f^*(w)$. Let $\psi_2 = \langle \cdot,f^*(w) \rangle \in V^*$ and $\theta = \psi_1 – \psi_2 \in V^*$.
Now consider ker$(\theta)$. If ker$(\theta)$ = $V$ then our claim is proven — namely $\psi_1 = \psi_2$.
Mistake: Firstly $\psi_1 \in W^*$, and not in $V^*$. And because $W \to W^*$ need not be an isomorphism the argument doens't follow.
EDIT: $\psi_1 \in V^*$ indeed. Namely $\psi_1: v \mapsto \langle f(v),w \rangle$ and $w \in W$ is fixed. However note that $\psi_1$ uses an inner product on $W$.
EDIT: Note that rk$(\theta) = 1$, so we have dim$\big($ker$(\theta)\big) = $ dim$(V) – 1$. Which fails my attempted proof.
Namely take dim$(V) = 1$. So that, ker$(\theta)$ = $\{0\}$. And thus for all $0 \neq v \in V$ we have $\psi_1 \neq \psi_2$.
Try 2: the same as Try 1 but with $v = x$.
I'm assuming that the argument can be proven by a less cumbersome method. Also I think that my method is not correct, since I seem to get stuck.
Suggestions or an alternative method are welcome.
EDIT: Here's a summary of Fischer's proof.
Define the linear map $\lambda: W \to V^*$ by $$\lambda(w) = \langle \cdot, w \rangle_W \circ f.$$ Next define the isomorphism $\sigma: V \to V^*$ by $$\sigma(v_0) = \langle \cdot, v_0 \rangle_V.$$ Now, consider the composition $$\varphi: \sigma^{-1} \circ \lambda: W \to V.$$ We then have $$\langle v, \varphi(w) \rangle_V = \langle f(v), w \rangle_W$$ for all $v \in V, w \in W$. We can now define $f^*:= \varphi$. This proves the theorem for the existence part.
For uniqueness, assume that $\psi$ satisfies $\langle f(v),w \rangle_W = \langle v, \psi(w) \rangle_V$. Fixing $w_0 \in W$, we get $$\langle v,\psi(w_0) – f^*(w_0) \rangle_V = 0$$ for all $v \in V$. Choosing $v = \psi(w_0) – f^*(w_0)$ we get $\psi = f^*$, which proves the uniqueness part.
Best Answer
Let's take a step back, and look at it from a more abstract perspective.
First, we define a map $\lambda \colon W \to V^\ast$ by
$$\lambda(w) \colon v \mapsto \langle f(v), w\rangle_W,$$
or $\lambda(w) = \langle\,\cdot\,,w\rangle_W \circ f$. Now, looking at it, we notice that $\lambda$ is linear (or, if the scalar field is $\mathbb{C}$, and the inner product linear in the first and antilinear in the second argument, $\lambda$ is antilinear; if the spaces are complex and the inner product is antilinear in the first, and linear in the second argument, consider $\langle w,f(v)\rangle_W$ instead):
$$\begin{align} \lambda(w_1+w_2)(v) &= \langle f(v), w_1 + w_2\rangle_W\\ &= \langle f(v), w_1\rangle_W + \langle f(v),w_2\rangle_W\\ &= \lambda(w_1)(v) + \lambda(w_2)(v)\\ &= \bigl(\lambda(w_1) + \lambda(w_2)\bigr)(v), \end{align}$$
and since $v$ was arbitrary, that means $\lambda(w_1+w_2) = \lambda(w_1) + \lambda(w_2)$. And for a scalar $s$, we have
$$\begin{align} \lambda(s\cdot w)(v) &= \langle f(v), s\cdot w\rangle_W\\ &= s\cdot \langle f(v), w\rangle_W\\ &= s\cdot \lambda(w)(v)\\ &= \bigl(s\cdot \lambda(w)\bigr)(v), \end{align}$$
that is, $\lambda(s\cdot w) = s\cdot \lambda(w)$ (for complex scalars, we would have to conjugate $s$ and $\lambda$ would be antilinear).
Now, with the (anti-) isomorphism $\sigma \colon V \to V^\ast$, given by $\sigma(v) = \langle\,\cdot\,,v\rangle_V$, i.e. $\sigma(v) \colon w \mapsto \langle w, v\rangle_V$ (or $\sigma(v) = \langle v,\,\cdot\,\rangle_V$, i.e. $\sigma(v) \colon w \mapsto \langle v,w\rangle_V$ if $V$ is a complex vector space and the inner product is antilinear in the first argument, as is customary in physics), consider the composition
$$\varphi = \sigma^{-1} \circ \lambda \colon W \to V.$$
Then $\varphi$ is linear (as the composition of two linear maps in the real case, and as the composition of two antilinear maps in the complex case), and for $v\in V,\; w \in W$, we have
$$\begin{align} \langle v, \varphi(w)\rangle_V &= \langle v, \sigma^{-1}(\lambda(w))\rangle_V\\ &= \lambda(w)(v)\tag{definition of $\sigma$}\\ &= \langle f(v),w\rangle_W,\tag{definition of $\lambda$} \end{align}$$
and that means we can define $f^\ast := \varphi = \sigma^{-1}\circ \lambda$.
It remains to be seen that there is only one such map $W \to V$.
Suppose that $\psi \colon W \to V$ is a map with $\langle f(v), w\rangle_W = \langle v, \psi(w)\rangle_V$ for all $v\in V,\; w\in W$.
Fix an arbitrary $w_0\in W$. Then for all $v\in V$
$$\begin{align} \langle v,\psi(w_0) - f^\ast(w_0)\rangle_V &= \langle v,\psi(w_0)\rangle_V - \langle v, f^\ast(w_0)\rangle_V\\ &= \langle f(v), w_0\rangle_W - \langle f(v), w_0\rangle_W\\ &= 0. \end{align}$$
In particular, we can choose $v = \psi(w_0) - f^\ast(w_0)$, so we have
$$\langle \psi(w_0) - f^\ast(w_0), \psi(w_0) - f^\ast(w_0)\rangle_V = 0,$$
which means $\psi(w_0) - f^\ast(w_0) = 0$. Since $w_0 \in W$ was arbitrary, it follows that $\psi = f^\ast$, and the uniqueness is also established.