Are there any other sequence of continuous functions like $x^n$ which converges pointwise to $0$ except at $x=1$ on$ [0,1]$?

Question

I was amazed at the beauty of the sequence of continuous functions $f_n(x)= x^n$. It certainly converges pointwise (not uniformly) on $[0,1]$. I want to know what are the characteristics of this particular sequence of continuous functions which give them a peculiar limit function which is continuous almost everywhere on $[0,1]$. How can we construct this type of sequence of functions? Certaily the sequence I want to construct should not be a subsequence of the given sequence and which is eventually different from the above sequence

Answer 1

Take $f_n(x) = (x^a)^n $ on $[0,1]$ , where $ a $ is any positive integer.

Take $f_n(x) =( \sin(\frac{πx}{2}) )^n $ on $[0,1]$

Also, take $f_n(x) = (e^{x-1})^n $ on $[0,1]$

Take $f_n(x) =( \tan(\frac{πx}{4}) )^n $ on $[0,1]$

Answer 2

Let $g(x)$ be any function such that $g(x)\in[0,1)$ for $x\in[0,1)$ and $g(1)=1$. For example,

$$g(x)=x^k$$ $$g(x)=sin\left({\pi x \over 2}\right)^k$$ $$g(x)={x^2+1\over 2}$$ $$g(x)=3x^2-2x^3$$ $$g(x)=6x^5-15x^4+10x^3$$

etc.

Then, the sequence

$$f_n(x) = g(x)^n$$

satisfies your requirements.

Answer 3

If you have $f_n(t)=t^n$,t$\in$[0,1] and you take a continuous g:[0,1]->$\mathbb R$ with g(1)=0(every function g with those conditions)then you can show that for every t$\in$[0,1] $(f_n.g)(t)$-->0 uniformly

Answer 4

I'll do you one better and work with a sequence with a pointwise limit that makes sense on all of $\mathbb{R}$. Consider the sequence of compactly supported smooth functions defined by

$$ f_n(x) = \exp(1)\cdot\exp\left(-\frac{1}{1-(n(1-x))^2}\right).$$

on $\left(1-\frac{1}{n},1+\frac{1}{n}\right)$ and $0$ elsewhere.

Answer 5

1. Take

$$f_n(x) = \frac{1}{1+n(1-x)}.$$

Then $f_n(x) \downarrow 0$ if $x\in[0,1)$ and $f_n(1)=1$. The following figure demonstrates the graph of $f_n$ on $[0,1]$ for $n=2^1,2^2,\dots,2^{14}$, colored in increasing order of frequency.

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2. Take

$$f_n(x) = \frac{2n}{1+(n(x-1)+\sqrt{2n-1})^2}.$$

Then $f_n(x) \to 0$ if $x \neq 1$ and $f_n(1)=1$. This example is interesting because $\sup_{x\in\mathbb{R}} f_n(x) = 2n$ diverges as $n\to\infty$. The following figure demonstrates the graph of $f_n$ on $[0,1]$ for $n=2^1,2^2,\dots,2^{14}$.