Consider a vector space $V$ which admits two different norms $\|\cdot \|_1$, $\|\cdot\|_2$ and is not complete with respect to either of them. Suppose moreover that the 1-norm is weaker than the 2-norm. If $\hat{V}_1$ and $\hat{V}_2$ denote the completions of $V$ with respect to the two norms, one concludes that $\hat{V}_2\subset \hat{V}_1$. As this inclusion indicates, there is a natural mapping of $\hat{V}_2$ to $\hat{V}_1$.
This appears on page 16 of Challifour's book on generalized functions where the author claims that the map thus defined is clearly linear and onto. It's not quite clear to me why the map is onto, i.e., why any Cauchy sequence with respect to the 1-norm is the image of some Cauchy sequence with respect to the 2-norm under our natural mapping.
Added in Edit As one of the examples below shows the claim in the book is incorrect. Unfortunately, on page 23 the author says that if the two norms are compatible, then the mapping described above is a bijection! (which is certainly wrong). The proof on page 23 however does not require bijectivity. A better exposition is given in Gelfand--Shilov page 14 where the authors clearly state in a footnote that the natural mapping need not be onto.
This is not true. Consider $V=C_c^\infty(\mathbb{R})$ (smooth functions with compact support) and $$\Vert f \Vert_1 := \left(\int_\mathbb{R}\vert f(x)\vert^2\right)^{1/2} \le \left(\int_\mathbb{R}\vert f(x)\vert^2\right)^{1/2} + \left(\int_\mathbb{R}\vert f'(x)\vert^2\right)^{1/2} =: \Vert f\Vert_2,$$ then $W^{1,2}\mathbb{R}\cong\hat V_2 \hookrightarrow \hat V_1 \cong L^2\mathbb{R}$ is the inclusion from a Sobolev space into the corresponding Lebesgue space, clearly not surjective.