Linear extension of “linear” map on dense subset of Banach space

Let X, Y be Banach spaces over the field \mathbb{F} (= \mathbb{R} \text{ or } \mathbb{C}).  Let Q be dense in \mathbb{F}, D be dense in X, such that D is closed under Q-linear combinations.  Let f \in Y^D, such that for all x, y \in D, p, q \in Q, f(px + qy) = pf(x) + qf(y) and for all x \in D, \| f(x) \| \le \| x \|. Then, there exists T \in L_1(X,Y) such that T|_D = f.

First, we define T:X \to Y.  Let x \in X. There exists (d_n) \subset D, such that d_n \to x.  Define T(x) := \lim_{n \to \infty} f( d_n ). We need to check that T is well-defined.  Let e_n \to x, e_n \in D.  Now, let \epsilon > 0.  Then, since (d_n) is Cauchy, there exists N such that n, m \ge N implies \| d_n - d_m \| < \epsilon.  Therefore, \| f(d_n - d_m) \| \le \| d_n - d_m \| < \epsilon, so (f(d_n)) converges in Y.  Now, there exists N such that n, m \ge N implies \| d_n - x \| < \epsilon, \| e_m - x \| < \epsilon. Therefore, \| e_m - d_n \| < 2 \epsilon, by the triangle inequality.  So \|f(e_m) - f(d_n) \| < 2 \epsilon. Hence
\| \lim_{m \to \infty} f( e_m ) - \lim_{n \to \infty} f( d_n ) \| \le 2 \epsilon.
Since \epsilon, x were arbitrary we see that T is well-defined on X.

Now, we check that T is linear.
Let x, y \in X.  Then, x = \lim_{n \to \infty} d_n, y = \lim_{n \to \infty} e_n,  where \{ d_n, e_n : n \in \mathbb{N} \} \subset D. Then, T( x + y ) = \lim_{n \to \infty} f( d_n + e_n ) = \lim_{n \to \infty} f( d_n ) + f( e_n ) = T(x) + T(y), so T is additive. The demonstration that T(rx) = rT(x), for r \in \mathbb{F} is quite similar.

Now, we need to show that T is bounded.  \| T(x) \| = \| \lim_{n \to \infty} f( d_n ) \| = \lim_{n \to \infty} \| f( d_n ) \| \le \lim_{n \to \infty} \| d_n \| = \| x \|. Thus T is bounded with \| T \| \le 1. \blacksquare

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