Once one of my lecturer said that if we worked in $L_2$, the complex inner product $$(u,v)=\displaystyle\int^L_0 u\ \bar v\ dx $$ works flawless for Fourier series, instead of working in $C^2$.
Question: I do't see the general picture that what is the problem with working in $C^2$. I am taking differential equations under Mathematical Methods in Physics I (in my Physics department), so there is almost no rigor in our lectures, I want to learn all the underlying idea Fourier series and this mentioned idea between $C^2$ and $L^2$. Any idea, comment, source, guidance, will be appreciated.
Your lecturer probably meant that the vector space $C^2([0,L])$, consisting of functions that are twice differentiable with continuous derivatives, is not complete if equipped with the scalar product $$ \langle f| g\rangle= \int_0^L f(x)\overline{g(x)}\, dx.$$ Among the various unpleasant consequences of this, there's the Fourier series fact that if you take an arbitrary series $$ S(x)=\sum_{k=-\infty}^\infty c_k e^{i\frac{2\pi}{L}kx} $$ then, even if $\sum_{k=-\infty}^\infty |c_k|^2<\infty $ (corresponding to the physical fact that $S$ has finite energy), you cannot conclude that $S\in C^2$. (In fact, it is not trivial to find conditions on $c_k$ that ensure $S\in C^2$. I don't think that a necessary and sufficient condition is known.)
On the other hand, the vector space $L^2([0, L])$, consisting of all measurable functions $f$ such that $\int_0^L |f(x)|^2\, dx<\infty$, is complete if equipped with the aforementioned scalar product $\langle\cdot|\cdot\rangle$. As a consequence of this, the finite energy Fourier series $S(x)$ always defines an element of $L^2([0, L])$. (Actually, all elements of $L^2([0, L])$ can be decomposed in a finite energy Fourier series.)
A good reference to be introduced to these things is the book by Stein and Shakarchi "Fourier analysis: an introduction", first volume.