Suppose I have a function $f:I \to \mathbb{R}$ that is continuous at $I$, except at a finite number of points, for example $ \{ C_n; \enspace (n \le K) \in \mathbb{N} \}$.

How can I build a sequence of continuous functions $f_n :I \to \mathbb{R}$ that converge pointwise to $f$?

My Idea, at first, was to take the Fourier series but I realized that it would not work because:

  • I don't know what kind of discontinuity points I have;
  • the Fourier series at discontinuity points converges to the mean value of the lateral limits, so we would not have pointwise convergence there.

1 Answer 1


Cut out a ball of radius $\tfrac1n$ around each discontinuity $x_k$ (assume $n$ is large enough that these balls are all disjoint). Then just define $f_n$ to agree with $f$ on the complement of the balls. On the ball around $x_k$, define $f_n$ to be linear so that it "spans the gap", i.e., connects the points $(x_k-\tfrac1n,f(x_k-\tfrac1n))$ and $(x_k+\tfrac1n,f(x_k+\tfrac1n))$.

It may be necessary to make two segments if the the linear part doesn't happen to agree with $f$ at $x_k$. But that's easy: first span the gap between $(x_k-\tfrac1n,f(x_k-\tfrac1n))$ and $(x_k,f(x_k))$, then span the gap between $(x_k,f(x_k))$ and $(x_k+\tfrac1n,f(x_k+\tfrac1n))$. This ensures that $f_n(x_k)=f(x_k)$ for all $n$ and $k$.

  • $\begingroup$ Yes, I think that is the idea, but when we take $n \to \infty$ the coefficient of our lines would have an indetermination. Because for the coefficient we would have something like $\frac{f(x_k +\frac{1}{n}) - f(x_k)}{1/n}$. How can we deal with this? $\endgroup$
    – H44S
    Nov 21, 2019 at 15:26
  • $\begingroup$ That doesn't matter. You asked for pointwise convergence, and this provides it. For any given $x\in\mathbb R$, it is true that $\lim\limits_{n\to\infty}f_n(x) = f(x)$ -- that's what pointwise convergence is. This is true whether or not it happens to be the case that $x$ is one of the $x_k$. $\endgroup$
    – MPW
    Nov 21, 2019 at 15:53

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