If $g \in L_2(\mathbb{R})$, then we can define the $L_2$ norm to have the following relationship: $\|g\|_2^2 = \int_\mathbb{R} g^2$. If $A\subseteq \mathbb{R}$, then we can define the norm of $L_2(A)$ to be $\|g\|_A = \left(\int_A g^2\right)^{1/2}$, or equivalently, one can define $\|g\|_A = \inf\{\|h\|_2: h\in L_2(\mathbb R), h_{|A} = g\}$.
My $\textbf{question}$ deals with when $L_2(A)$ is a Hilbert space with inner product $\langle g,h\rangle_A = \int_A g\bar{h}$. Is there an equivalent definition like they use for the norm? That is to say something like,
$ \begin{align*} \langle g,h \rangle_A = \inf \left\{\int_{\mathbb{R}} g^{'}h^{'} : g^{'},h^{'}\in L_2(\mathbb R), g^{'}_{|A} = g, h^{'}_{|A} = h\right\} \end{align*} $
I am a bit uncertain because of the inner product is not non-negative and so I am not sure if this definition holds.
Suppose $H$ to be a Hilbert space, $X$ normed linear space. Assume that there is an injection $$ I\in \mathcal L(X,H). $$ Then one can define an inner product on $X$ $$ \langle x,y\rangle_X := \langle Ix,Iy\rangle_H. $$ The space $X$ equipped with the norm $\|x\|_I:=\sqrt{\langle x,y\rangle_X}$ is pre-Hilbert space.
In the case of $H=L^2(\mathbb R)$, $X=L^2(A)$, define $I$ by $$ x\in L^2(A) \mapsto (Ix)(t):=\chi_A(t)x(t), $$ so $I$ extends $x$ outside of $A$ by $0$. In this case, the usual inner product on $L^2(A)$ coincides with the inner product induced by the injection.
The construction of the inner product on $L^2(A)$ with infimum as in the question does not work: the infimum is always $-\infty$.