# Does the monotone pointwise limit of continuous functions has left and right limit everywhere?

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?

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

$$f_n(x)=d_{(\frac{1}{k+1},0,\frac{\frac{1}{k}+\frac{1}{k+1}}{2},1)}(x)+S_n.$$

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

$$f_n(x)=d_{(\frac{\frac{1}{k}+\frac{1}{k+1}}{2},1,\frac{1}{k},0)}(x)+S_n.$$

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.

• 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. Commented May 3, 2021 at 18:53
• 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 :) Commented May 3, 2021 at 20:36