Show that for all $\epsilon>0$ exists a continous function $g$ such that $\|g-f\|_2<\epsilon$ Suppose that $f$ is a continous function at $[a,b]$ except for a jump discontinuity at $c\in(a,b)$. Show that for all $\epsilon>0$ exists a continous function $g$ such that $\|g-f\|_2<\epsilon$
My try:
We just need to show that $\|g-f\|_2<\epsilon_0$ for one jump discontinuity and then add the error of the finite number of them in orther to get $\|g-f\|_2<\sum_{i=0}^{n}\epsilon_i=\epsilon$.
Let $S_n(f)$ be the Fourier series of $f$
Now let $N>m$ such that $\|S_N(f)-f\|_2<\epsilon_0$.
Then the sum of all the jump dicontinuities must be $$\|S_N(f)-f\|_2 = \|g-f\|_2<\sum_{i=0}^{n}\epsilon_i=\epsilon$$
Any suggestions would be great!
 A: First, reduce our consideration to a step function as follows. Let $c_1,\dots,c_n$ denote the values of $x \in (a,b)$ at which $f$ has a jump discontinuity.
Define $h_j(x)$ by
$$
h_j(x) = \begin{cases}
0 & x < c_j\\
f(c_j) - \lim_{x \to c_j^-}f(x) & x = c_j\\
\lim_{x \to c_j^+}f(x) - \lim_{x \to c_j^-} f(x) & x > c_j
\end{cases}
\qquad j = 1,\dots,n.
$$
Define $\phi(x) = f(x) - (h_1(x) + \cdots + h_n(x))$. We see that $\phi(x)$ is continuous. Now, suppose that $\gamma_1,\dots,\gamma_n$ are continuous functions with $\|\gamma_j - h_j\|_2 < \epsilon/n$ for each $j = 1,\dots,n$. If we define
$$
g(x) = \phi(x) + \gamma_1(x) + \cdots + \gamma_n(x), 
$$
then we find that
$$
\begin{align}
\|g - f\|_2 &=
[\|\phi + \gamma_1 + \cdots + \gamma_n] - [\phi + h_1 + \cdots + h_n]\|_2
\\ & = \|(\gamma_1 - h_1) + \cdots + (\gamma_n - h_n)\|_2
\\ & \leq \|\gamma_1 - h_1\|_2 + \cdots + \|\gamma_n - h_n\|_2 < n \cdot (\epsilon/n) = \epsilon.
\end{align}
$$
That is, it suffices to consider the problem in which we have
$$
f(x) = \begin{cases}
0 & x< c\\
k_1 & x = c\\
k_2 & x > c.
\end{cases}
$$
One choice of $g$ that works well is a piecewise-linear function that satisfies $g(x) = f(x)$ for all $x$ with $|x - c| > \delta$, for some $\delta > 0$ (that depends on $k_2$).
