Why periodic functions form a dense subset in $ C[a,b]$ with $L^2$ norm?

Question

Let's consider the linear space $C[a,b]$ but with $L^2$ norm $$ \|f\|=(\int_a^b |f(t)|^2dt)^{\frac{1}{2}} $$

How to prove that the subspace $$ V=\{f\in C[a,b]: f(a)=f(b)\} $$ is dense in this normed space.

Thanks

Show that $V_0 = { f \in C[a,b] : f(a) = f(b) = 0}$ is dense. — Daniel Fischer, Jun 27 '15 at 15:54

score 2 · Accepted Answer · answered Jun 27 '15 at 17:34

Let $f \in C[a,b].$ Set

$$f_n(x) = f(x)[(x-a)(b-x)]^{1/n}, n = 1,2,\dots .$$

Then $f_n \in V, n= 1,2,\dots$ and $f_n \to f$ uniformly on $[a+\delta, b-\delta]$ for all small $\delta>0.$ This and the uniform boundedness of the $f_n$ will give $f_n\to f$ in the $L^2$-norm.

Why periodic functions form a dense subset in $ C[a,b]$ with $L^2$ norm?

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