Prove that there exists a linear functional $l$ definied on $H$ that is not bounded . 
Let $H$ be an infinite-dimensional Hilbert space . Then there exists a linear functional $l$ defined on $H$ that is not bounded .  

My attempt :
Let $B=\{e_{\alpha}\}_{\alpha \in I}$ denote the basis of $H$ , then we can construct a countable subset of $B$ . WLOG , we say $A=\{e_1,...,e_n,...\}$ is an orthonormal subset of $B$ . Then construct functional $l$ as follows :
$$l(e_k)=k \,\,\,\,\,\,\,\,\,\text{whenever $k \in N^*$  }$$
$$l(f)=0 \,\,\,\,\,\,\,\,\,\text{whenever $f \in B-A$  }$$ 
Then if $l$ is well-defined on $H$ then $||l(e_k)||=k$ is not a bounded functional .  
My qusetion :
I want to show if $I' \subset I$ is countable and $$f=\sum_{\alpha \in I'} a_{\alpha}e_{\alpha}$$ then $$l(f)=\sum_{\alpha \in I'} p(\alpha)a_{\alpha} \,\,\,\,\,\,\,\,\,\,\,\,\,\text{here $p(\alpha)=n $ whenever $e_{\alpha}=e_n$ and $p(\alpha)=0$ otherwise}$$
However , this might leads some contradiction . Consider $f_1=\sum_{n=1}^{\infty} \frac1n e_n$ , it is obvious $f_1 \in H$ but $l(f_1)=\sum_{n=1}^{\infty} 1$ is not well-defined .  
 A: This actually does not work in quite the way you are trying to make it work. The reason is that there aren't any "constructible" discontinuous linear functionals on complete normed vector spaces. 
There is a reasonable way to try to proceed, which is basically what you tried to do but cleaned up a bit. Take a Schauder basis $B$ of $H$, and take a countable subset $C$ with enumeration $e_k$. Define $l$ on the span of $C$ by defining $l(e_k)=k$ and extending by linearity. Define $l$ to be $0$ on the span of $B \setminus C$. Then extend by linearity again, hopefully extending $l$ to all of $H$.
Except you didn't actually extend $l$ to all of $H$: you only extended $l$ to the direct sum of the span of $C$ and the span of $B \setminus C$. Because $B$ is just a Schauder basis, this direct sum is in general not all of $H$; in fact it does not even necessarily contain the closed span of $C$. To get an extension to all of $H$, you would need to be sure that for all $f \in H$, if the projection of $f$ onto the span of $C$ is $\sum_{k=1}^\infty a_k e_k$, then $\sum_{k=1}^\infty k a_k$ (which is what you would expect $l(f)$ to be) converges. In fact it must converge absolutely as well. This property just doesn't hold (your example shows why), so $l$ is not defined on all of $H$.
You can circumvent this issue by taking $B$ to be a Hamel basis in the first place. Then the argument I gave works, because the direct sum of the span of $C$ and of the span of $B \setminus C$ is just $H$. But a Hamel basis of an infinite dimensional normed vector space requires some kind of choice principle to "construct" it.
