In Real analysis by Folland, the Plancherel Theorem is as follows:
If $f \in L^1 \cap L^2$, then $\hat{f}\in L^2$, and $\mathcal{F}|(L^1 \cap L^2)$ extends uniquely to a unitary isomorphism on $L^2$.
In the proof it defines $\mathcal{X}=\{f\in L^1:\hat{f}\in L^1\}$. Then it proves that $\mathcal{F}$ preserves the inner product on $\mathcal{X}$ and $\mathcal{F}(\mathcal{X})=\mathcal{X}$ is isomorphism and also $\mathcal{X}\subset L^2$ and $\mathcal{X}$ is dense in $L^2$.
Then it says "$\mathcal{F}|\mathcal{X}$ extends by continuity to a unitary isomorphism on $L^2$".
My question is that "what it means by continuity" and "how can we extend" and "why the extension is unique"?
Let $M$ be a dense linear subspace of a Hilbert space $H$ and $T:M \to H$ be a linear map that preserves inner products. If $x \in H$ then there is a sequence $\{x_n\} \in M$ converging to $x$. Since $\|Tx_n-Tx_m\|=\|x_n-x_m\| \to 0$ it follows that $\{Tx_n\}$ is Cauchy sequence . Hence it converges in $H$. Let $Tx=lim TX_n$. This gives the extension of $T$ to $H$. From the continuity of inner product you can verify that the extended map also preserves inner products.
Uniqueness is obvious. If two continuous functions coincide on a dense subset they coincide everywhere.