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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!

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    $\begingroup$ "tal que" = "such that" $\endgroup$ Nov 12 '20 at 20:22
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    $\begingroup$ You say "a finite amount of jump discontinuity at one value $c \in (a,b)$". Does that mean that each of finitely many jump discontinuities occur between $a$ and $b$, or does it mean that there is exactly one value $c \in (a,b)$ at which $f$ has a jump discontinuity? $\endgroup$ Nov 12 '20 at 20:28
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    $\begingroup$ If you mean the latter, then what do you mean by "all the jump discontinuities"? $\endgroup$ Nov 12 '20 at 20:33
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    $\begingroup$ So in other words, my first statement was correct: each of finitely many jump discontinuities occur at a value between $a$ and $b$. $\endgroup$ Nov 12 '20 at 20:49
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    $\begingroup$ When you say "$f$ is a continous function at $[a,b]$ except for a jump discontinuity at..." then the meaning is that there is (overall) one jump discontinuity for the function $f$. I suspect that this is a linguistic difference between English and Spanish. $\endgroup$ Nov 12 '20 at 20:51
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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 $k_2$).

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