Demonstration of the completness of an orthonormal set of functions I find this concept of completness a little bit dense when it comes to prove this property of some set of orthonormal functions. In one of my classes, my professor proved this for the orthonormal set of functions $\left\{ \sqrt{2/L} \sin( n \pi x/L) \right\}$, but it did not convince me, even though I can't tell if there is something wrong mathematically speaking. He parted from the very condition of completness, i.e.,
$$\sum_n \frac{2}{L}\sin(\frac{n\pi}{L}x')\sin(\frac{n\pi}{L}x)=\delta(x-x')\;\;\;\;\;\;\;\;\;\;\;(1)$$
and he supposed that, being the set a complete one, then one can describe any funcion in terms of such set. He then wrote that
$$\delta(x-x')=\sum_nC_n\frac{2}{L}\sin(\frac{n\pi}{L}x)\;\;\;\;\;\;\;\;\;\;\;(2)$$
Then, taking advantage of the orthogonality of the set, on the interval $0\leq x\leq L$, from the equation (2)
$$\int_{0}^{L}\delta(x-x')\frac{2}{L}\sin\left(\frac{m\pi}{L}x\right)\mathrm{d}x=\sum_nC_m \delta_{m,n}=C_m$$
$$\therefore \frac{2}{L}\sin\left(\frac{m\pi}{L}x'\right)=C_m\;\;\;\;\;\;\;\;\;\;\;(3)$$
and replacing (3) into (2), then one gets the condition for completeness in (1). Even if this is correct, I can't tell why. Also, I would like to know how the proof for completness would be carried out taking the same condition but in Dirac's notation, that is, $\sum_n |\phi_n><\phi_n|=1$, but I have no idea how to proceed.
 A: Sorry, I really don't under what the question is in the first part.
For the second part, let
$$|v>=\sum_n c_n |\phi_n> \tag{1} $$
where  $|v>$ is a vector (ket) in a finite dimensional space with an orthonormal basis $|\phi_n>$ and 
$$<v|=\sum_n c^*_n <\phi_n|$$
is an adjoint or dual vector (bra).
Orthonormality of the basis for the kets and bras gives $<\phi_m|\phi_n>=\delta_{m,n}$.
From the decomposition of $v$ in equation $(1)$,
$$c_n=<\phi_n|v>$$.
Noting that $c_n$ is a scalar, substitute the value of $c_n$ back into equation $(1)$ and re-arrange
$$|v>=\sum_n c_n |\phi_n>=\sum_n <\phi_n|v>|\phi_n>=\sum_n |\phi_n><\phi_n|v>.$$
Breaking apart $<\phi_n|v>=<\phi_n|1|v>$ (you can always insert a $1$ between a bra and a ket) yields
$$|v>=\sum_n {|\phi_n><\phi_n|}1|v>=\sum_n {|\phi_n><\phi_n}|v>$$
which implies
$$\sum_n |\phi_n><\phi_n|=1.$$ 
I'm using the $1$ (or the identity) in $1|v>$ for clarity - it's typically not necessary.  
A: Do you accept that known complete sets of functions, I.e. delta functions, exist?
If so, proving that you alternate set can reproduce the delta functions is enough.   Since combining functions is linear, you can thereby always use your functions to make the deltas you need to make anything else. 
