# 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!

• "tal que" = "such that" Nov 12 '20 at 20:22
• 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? Nov 12 '20 at 20:28
• If you mean the latter, then what do you mean by "all the jump discontinuities"? Nov 12 '20 at 20:33
• So in other words, my first statement was correct: each of finitely many jump discontinuities occur at a value between $a$ and $b$. Nov 12 '20 at 20:49
• 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. Nov 12 '20 at 20:51

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 $$k_2$$).