$X$ is a normed space. $(X_n)_n$ is a sequence of finite dimensional subspaces of $X$ and $X_n\subset X_{n+1}$ and $X$ is the closure of the union of all $X_n$.
Show that there exists a sequence of linear functionals $f_n\in X^{*}$ such that $\left\Vert f_n \right\Vert=1$ and $f_n(x)=0$ for all $x\in X_n$. Show also that $f_n(z)$ converges to $0$ for any $z\in X$.
I defined $f_n:X\rightarrow \mathbb{R}$ where $f_n(x)=0$ if $x\in X_n$ and $f_n(x)=1$ if $x\notin X_n$. The norm equals to 1 is quite easy but I have problem showing that my $f_n$ is linear by construction.
Because if $x,y\notin X_n$, then I don't have $f_n(x+y)=f_n(x)+f_n(y)$ since the LHS is 1 and the RHS is 2 and thus I cannot claim that my $f_n\in X^{*}$.
Any advice? How to show the last part converge to $0$? Is my construction of $f_n$ correct?
$X \neq X_n$ so there exists some $y \in X \setminus X_n$. Then each $X_n$ is finite dimensional, and hence is a closed convex space. In addition $\{y\}$ is a compact convex space, moreover $\{y\}$ and $X_n$ are disjoint. So by the Hahn-Banach seperation theorem (derived from the Hahn-Banach theorem, sometimes also called a geometric version) we can seperate $X_n$ and $\{y\}$ strictly by a hyperplane. In other words there exists $f_n \in X^*$ and $\alpha \in \mathbb{R}$ such that $$ f_n(x) < \alpha < f_n(y) \quad \forall x \in X_n $$ But then $f_n(\lambda x) = \lambda f_n(x) < \alpha$ for all $\lambda \in \mathbb {R}, x \in X_n$ so we must in fact $f_n(x) = 0$ for all $x \in X_n$. Moreover $f_n(y) > 0$ so $f_n \neq 0$ therefore we can rescale $f_n$ such that $\lVert f_n \rVert = 1$.
Now let $z \in X = \overline{\cup_{n=1}^{\infty} X_n}$. Then $z = \lim_{i \to \infty} x_i$ for some $(x_i)_{i=1}^{\infty} \subset \cup_{n=1}^{\infty} X_n$. It should be clear for any $x \in \cup_{n=0}^{\infty} X_n$ that $f_n (x) \to 0$ as $n \to \infty$, we use this along with uniform boundedness of $(f_n)$ to show $f_n (v) \to \infty$.
Given $\epsilon > 0$, $v = \lim_{i \to \infty} x_i$ therefore $\exists I$ such that $\lVert v - x_I \rVert < \epsilon$. Then $f_n (x_I) \to 0$ as $n \to \infty$ therefore there exists $N$ such that for all $n \geq N$ we have $\lVert f_n(x_I) \rVert < \epsilon$.
Then for all $n \geq N$ we have $$ \lVert f_n (v) \rVert \leq \lVert f_n (v) - f_n (x_I) \rVert + \lVert f_n (x_I) \rVert \leq \lVert f_n \rVert \lVert v - x_I \rVert + \lVert f_n (x_I) \rVert < 2 \epsilon $$ Hence $f_n (v) \to 0$ as $n \to \infty$ as required.