# Is there a sequence of continuous functions which converges nowhere

find $$\{f_n \}$$ a sequence of continuous functions on $$\mathbb{R}$$ such that $$0 \leq f_n(x) \leq 1$$ and $$\lim_{n \rightarrow \infty} \int_0^1 f_n(x)dx =0$$, but $$f_n(x)$$ does not converge anywhere on $$[0,1]$$.

Everytime I am thinking with a sequence, I come with one satisfies some of the conditions but violate one condition. For example, $$f_n(x)=x^n$$ is continuous sequence, and $$f_n \rightarrow 0$$ so limit the integral is zero, but it is pointwise convergent. , I tried to create a function similar to Dirchlet function, as they not converge pointwise, but they converge on part of the domain and they are not continuous.

• You start with a conditional, “If”, but there is never a conclusion (“then”). The sentence is not grammatical and you should really say what you mean. I think “Does there exist...” rather than “If”... Mar 23, 2021 at 23:41
• I updated the question! Mar 23, 2021 at 23:56

Rather than give formulas, I will describe the graphs of the functions, which are piecewise linear. Think of a “discretely moving wave”, where after the wave is done moving, we switch to a “thinner” wave.

Let $$f_1$$ go from $$(0,1)$$ to $$(\frac{1}{2},0)$$ to $$(1,0)$$. Let $$f_2$$ go from $$(0,0)$$ to $$\frac{1}{2},1)$$ to $$(1,0)$$. Let $$f_3$$ go from $$(0,0)$$ to $$\frac{1}{2},0)$$ to $$(1,1)$$.

Then let $$f_4$$ go from $$(0,1)$$ to $$(\frac{1}{4},0)$$ to $$(1,0)$$. Let $$f_5$$ go from $$(0,0)$$ to $$(\frac{1}{4},1)$$ to $$(\frac{1}{2},0)$$ to $$(1,0)$$. Then $$f_6$$ from $$(0,0)$$ to $$(\frac{1}{4},0)$$ to $$(\frac{1}{2},1)$$ to $$(\frac{3}{4},0)$$ to $$(1,0)$$. Then $$f_7$$ go from $$(0,0)$$ to $$(\frac{1}{2},0)$$ to $$(\frac{3}{4},1)$$ to $$(1,0)$$. Then $$f_8$$ from $$(0,0)$$ to $$(\frac{3}{4},0)$$ to $$(1,1)$$.

Next do the same thing with the partition $$0\lt \frac{1}{8}\lt\frac{1}{4}\lt\frac{3}{8}\lt \frac{1}{2}\lt \frac{5}{8}\lt\frac{3}{4}\lt\frac{7}{8}\lt 1$$. Then partitioning $$[0,1]$$ into $$16$$ equal subintervals, etc.

Each new batch of functions has a smaller integral, all positive, but converging to $$0$$. The functions are all continuous. And $$f_n(x)$$ does not converge for any $$x$$ because you can always find arbitrarily large values of $$n$$ where $$f_n(x)=0$$ and values where $$f_n(x)$$ is very close to $$1$$ (by approximating $$x$$ with a rational with denominator a power of $$2$$).

• I am curious to ask if these functions have a certain pattern or compact formula!! Mar 23, 2021 at 23:58
• It is really a clever example! Mar 23, 2021 at 23:58
• @math_for_ever: I’m sure we could come up with a relatively succint formula in terms of the nearest power of $2$ or something, but I don’t think there’s much point in expending the effort... Mar 23, 2021 at 23:59
• you are right! Does this sequence has a name? Mar 24, 2021 at 0:11
• @fourelements the subsequence consisting of the very first wave of each size converges pointwise to $\chi_{\{x=0\}}$.
– Alan
Jul 11, 2021 at 23:53