Let $X,Y$ be normed vector spaces. Suppose we have already defined a bilinear form $$ X \times Y \rightarrow k$$ that satisfies $$|\langle x,y \rangle| \le |x|_X |y|_Y$$ and $k$ is complete scalar field $\Bbb R$ or $\Bbb C$.
Now let us take the completion of $X$ and $Y$ wrt to their norms. Can we extend this to a bilinear form on the completion?
I think so. This is my argument:
Let us take the completion where an element is an equivalence class Cauchy sequence modulo sequences convergent to $0$ in the original norm. Let $[(x_n)], [(y_n)]$ be two elements in completions, $X',Y'$.
Define, $$ \langle [(x_n)], [(y_n)] \rangle := \lim_{n} \langle x_n, y_n \rangle $$ This is well defined: The sequence on RHS is Cauchy, since $$ |\langle x_n, y_n \rangle - \langle x_m, y_m \rangle | \le |\langle x_n - x_m, y_n \rangle | + |\langle x_m, y_n - y_m | \le |x_n - x_m|_X |y_n|_Y + |x_m|_X |y_n- y_m|_Y $$ using original condition. So the limit of RHS exists.
Also this is independent of the equivalence class chosen, since the same equation shows that limit if $x_n$ is a Cauchy sequence which limits to $0$, then the limit of the inner product is $0$.
The inequality still holds: Taking norms, since limits commute with it, we have $$ | \lim \langle x_n, y_n \rangle | \le \lim |x_n|_x |y_n|_Y = |[x_n]|_{X'} |[y_n]|_{Y'}$$ by the definition of the norm in the completion.
Is the statement true? and if so why (and possibly comment if my argument is correct).