Let
$\mathcal{H} = \{f:X \rightarrow \mathbb{R} : \exists w \in \mathbb{R^d}, f(x) = \langle w,x \rangle_{\mathbb{R}^d}, \forall x \in X \} $
where $\langle \cdot , \cdot \rangle$ denotes inner product.
How do you prove that $ w \rightarrow f$ is bijective?
I know that this might be a very simple question for people with more math experience than me, but have patience, I don't know how to do this.
I did try proof by contradiction, assume that there exists $f'(x) = \langle w', x\rangle$ and $f(x) = \langle w, x\rangle$ such that $f(x) = f'(x)$ but $w \neq w'$. But I just got some empty equations that didn't lead me anywhere:
$f'(x) = \langle w', x\rangle$
$f(x) = \langle w, x\rangle$
$\langle f, f'\rangle = \langle w, w'\rangle$
$\langle f(x), f'(x)\rangle = \langle \langle w', x\rangle, \langle w, x\rangle\rangle$
I do know that the inner product is symmetric, linear and positive-definite. Not sure if that helps though... any ideas?
I think I even came up with a counter example for the special case of the "normal" dot product:
$\langle0.25, 0.75 \rangle \cdot \langle 2, 2 \rangle = \langle 0.75, 0.25 \rangle \cdot \langle 2, 2 \rangle $
but $\langle0.25, 0.75 \rangle \neq \langle 0.75, 0.25 \rangle$
Why is my counter example invalid?
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I am also not sure what is an appropriate tag for this Question.
If $f(x) = f'(x)$ i.e.$\langle \omega, x\rangle = \langle \omega', x\rangle$ for all $x$, then $$\langle \omega - \omega', x\rangle = 0$$ for all $x$.
Then take $x= \omega - \omega'$, we get $$\langle \omega - \omega', \omega - \omega'\rangle = 0$$
which means $\omega - \omega' = 0$, i.e. $\omega = \omega'$