Weak topology on unit ball I have the following problem:
$X \in \mathbb{K}$ linear space and $F,f_1,\ldots,f_n : X \rightarrow \mathbb{K}$ linear with
$$ \bigcap_{i=1}^n N (f_i) \subset N (F)$$
Show that $F$ is a linear combination of $f_i$.
Maybe it works with induction but I have no clue how to start.
This should help me to solve then why $B^\circ _1 (0) = (x \in X : ||x||_x <1)$ (unit ball) is empty for weak topology on $X$ with $(X,||.||_x)$ being an infinite dimensional normed space. 
Can someone help me to show that? especially with the linear combination?
 A: $\newcommand{\K}{\mathbb K}$ 
As stated the problem really doesn't make sense. I'm going to assume you meant to write $$\bigcap_{i=1}^nN(f_i)\subset N(F).$$
Define $T:X\to\K^n$ by $$Tx=(f_1x,\dots,f_nx).$$
Let $Y=T(X)$, the image of $Y$ under $T$. Attempt to define $L:Y\to\K$ by $$L(Tx)=F(x)\quad(x\in X).$$
That defines a linear map from $Y$ to $\K$ if it is well-defined. That is, we need to show that 


If $Tx=Tx'$ then $Fx=Fx'$.


But this is clear: If $Tx=Tx'$ then $x-x'\in N(f_j)$ for every $j$, hence $x-x'\in N(F)$, so $Fx=Fx'$.
So $L:Y\to\K$ is linear. We can extend $L$ to a linear map $L:\K^n\to\K$. (If this is not clear, start with a basis $B$ for $Y$. Since $B$ is an independent subset of $\K^n$ there exists a basis $B'$ for $\K^n$ with $B\subset B'$. Now extend $L$ from $Y$ to $\K^n$ by declaring $Lb=0$ for every $b\in B'\setminus B$.)
Now since $L$ is a linear functional on $\K^n$ there exist $c_1,\dots,c_n\in\K$ such that $$Lk=c_1k_1+\dots c_nk_n\quad(k\in\K^n).$$
Hence if $x\in X$ then $$Fx=L(Tx)=c_1f_1(x)+\dots+c_nf_nx.$$
Edit Note the result is false for infinitely many $f_j$. for example let $X=\ell_1$, $f_jx=x_j$, $Fx=\sum_{j=1}^\infty x_j$.
