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In Functional Analysis, Hahn - Banach Theorem can be stated as follows:

" Let $X$ be a real or complex vector spave and $p$ a real-valued functional on X which is additive, that is, for all $x$, $y$ in $X$,

$$p(x+y) \le p(x) + p(y),\tag1\label1$$

and for every scalar $\alpha$ satisfies

$$p(\alpha x) = |\alpha| p (x).\tag2\label2$$

Furthermore, let $f$ be a linear functional which is defined on a subspace $Z$ of $X$ and satisfies

$$|f(x)| \le p(x)\quad \text{for all } x \in Z.\tag3\label3$$

Then $f$ has a linear extension $\widetilde{f}$ from $Z$ to $X$ satisfying

$$|\widetilde{f}(x)| \le p(x)\quad \text{for all } x \in X.\tag{3*}\label{3*}"$$

Introductory to Functional Analysis with Applications, Erwin Kreyszig, Theorem 4.3 -1, p291.

My question is the following:

Suppose that $\{f_n\}$ is a convergent sequence of linear functionals under the uniformly convergence in the subspace $Z$ of $X$ satisfying all conditions \eqref{1}, \eqref{2}, \eqref{3}. Can we extend $\{f_n\}$ to become another convergent sequence in the whole space $X$, which is still satisfying \eqref{3*}?

Thank you so much for your attention.

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Trần Linh
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    what do you mean by uniformly convergent? $Z$ is a Subspace of $X$ which isn't endowed with a norm so far... – Nathanael Skrepek Mar 14 '17 at 14:44
  • Thank you for your comment. By uniform convergent, I mean $| f_n (x) - f (x) | \rightarrow 0$ for every $x \in Z$ – Trần Linh Mar 14 '17 at 14:49
  • Taking, for example, $Z = {0}$, I think it is clear that the answer is no. – Roberto Rastapopoulos Mar 14 '17 at 14:51
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    that is pointwise convergence. if you mean $\sup_{x\in Z}|f_n(x)-f(x)| \to 0$ i am afraid to say that this is impossible for $Z \neq {0}$. – Nathanael Skrepek Mar 14 '17 at 14:54
  • Thank you for your comment. Actually, it is not the norm that I expected but it is also useful. Could you please explain in more detail? On book "Introductory Functional Analysis with Applications" (which I am using), they define as the following "Let $X$ and $Y$ be normed spaces. A sequence $(T_n)$ of operators $T_n \ in B(X,Y)$ is said to be uniformly operator convergent if $(T_n)$ converges in the norm on $B(X,Y)$. – Trần Linh Mar 14 '17 at 15:06

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