Consider a linear map $f : V \to W$ of Euclidean vector spaces. If $V$ is finite-dimensional, we can construct a unique linear function $f' : W \to V$ such that $\forall v \in V \ \forall w \in W : \left<f(v), w\right> = \left<v, f'(w)\right>$.
If we only assume $W$ is finite-dimensional instead, the usual construction via dual maps doesn't work (we need an orthonormal basis for $W$). However, I'm not sure how to prove that there is no such map $f'$ at all. I was told as a hint that one could use the condition $\left<f(v), w\right> = \left<v, f'(w)\right>$ to arrive at a contradiction. I don't know whether I'm supposed to directly arrive at a contradiction by assuming that $V$ is not finite-dimensional or somehow construct a counterexample. I tried to figure out in which cases the existence of an adjoint fails but I've only arrived at dead ends.
Any help would be appreciated.
I have constructed the following counterexample:
We consider $V = C([0, 1], \mathbb{R})$ with the usual scalar product and $W = \mathbb{R}$ with the standard scalar product. Further, let $T : V \to W$ be given by $T(f) = f(0)$. Suppose there was a linear mapping $T' : W \to V$ with $\forall f \in V \ \forall \alpha \in W : \left< T f, \alpha \right> = \left< f, T' \alpha \right>$. Since $T'$ must be a linear mapping, $\alpha$ can be extracted. If $\alpha \neq 0$, it then follows that $$ \forall f \in V \ \forall 0 \neq \alpha \in W : f(0) = \int _{0}^{1} f(x) T'(1) \; \mathrm{d}x . $$ We write $g(x) = T'(1)$.
The sequence of functions $(f_{n})_{n} = (1 - x)^{n}$ converges pointwise to $f = 0$, but $f_{n}(0) = 1$ for all $n$. Moreover, all $f_{n}$ are continuous. Since $g$ is Riemann integrable and $|f_{n} g| \leq |g|$ holds, the dominated convergence theorem says that $$ \underbrace{ \lim_{ n \to \infty } \underbrace{ \int _{0}^{1} f_{n}(x) g(x) \; \mathrm{d}x }_{ = f_{n}(0) = 1 } }_{ = 1 } = \underbrace{ \int _{0}^{1} \underbrace{ f(x) }_{ = 0 }g(x) \; \mathrm{d}x }_{ = \int _{0}^{1} 0 \; \mathrm{d}x = 0 }, $$ which is a contradiction. So no such $g$ exists and no such $T'$ exists.
Is this correct?