Let $\{f_n(x)\}_{n\geq 0}$ be a sequence of continuous functions such that $f_n(x) > f_{n + 1}(x)$ for all $n\geq 0$. Assume that $f_n(x)$ converges pointwise to a function $f(x)$. It is clear that $f$ is not necesarilly continuous, but, does it have right and left limit everywhere?


1 Answer 1


Here is an example of such a sequence defined on $[0,1]$, where the limit as $x \to 0$ does not exist. To simplify notation, and not calculate all the lines I will use explicitly, I will denote by $d_{(x_1,y_1,x_2,y_2)}(x)$ the equation for the line passing through $(x_1,y_1)$ and $(x_2,y_2)$.

Choose your favorite converging series with only positive terms. I will write it down as $S=\sum_{k=1}^\infty a_n$. Let $S_n=\sum_{k=1}^n a_n$ be the sequence of partial sums associated to this series.

For a given integer $n$, $f_n(x)$ will be $S_n$ if $x \in [0,\frac{1}{n+1}]$. If $x \in (\frac{1}{k+1},\frac{\frac{1}{k}+\frac{1}{k+1}}{2}]$ for some $k=1,2,...,n$, we let


If $x \in (\frac{\frac{1}{k}+\frac{1}{k+1}}{2},\frac{1}{k}]$ for some $k=1,2,..,n$, we let


This defines our function. I will let you verify that it respects all your criteria.

Denote by $f$ the pointwise limit of $f_n$. Notice that for all positive integers $j$, we have that $f(\frac{1}{j})=S$ and $f(\frac{\frac{1}{j}+\frac{1}{j+1}}{2})=1+S$. This gives us two sequences that tend to $0$ with $f$ having a different value for each of the sequences, and so the limit as $x$ goes to $0$ of $f$ does not exist!

If anything needs clarification, let me know! This was a fun problem to work on, thank you for posting the question.

  • $\begingroup$ Ah, I just noticed that I inverted the inequality you were using and instead supposed $f_n(x)<f_{n+1}(x)$. Simply consider $g_n=-f_n$ to have the desired inequality, and the rest works out the same. $\endgroup$
    – Jonah
    Commented May 3, 2021 at 18:53
  • $\begingroup$ Great counterexample! So it isn't true... I was hoping it was... pointwise convergence is so weak when it comes to inheriting properties :( I'm glad you had a good time working on this... thanks for helping :) $\endgroup$
    – Aguazz
    Commented May 3, 2021 at 20:36

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