Existence theorem about the adjoint.

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.