Suppose $f_n(x)=x^n - x^{2n}$ , $x \in [0,1]$. Dose the sequence of functions $\lbrace f_n \rbrace$ converge uniformly?

I try to show that the sequence is Cauchy. But I get stuck.

Suppose $n>m$. Then $|f_n(x)-f_m(x)|=|x^n-x^{2n}-(x^m-x^{2m})| \leq |x^n -x^m|+|x^{2n}-x^{2m}|=|x^n(1-x^{m-n})|+|x^{2n}(1-x^{2(m-n)})|$

After that, I try to use definition of uniform convergence. Clearly the sequence of functions converges pointwise to $0$.

$|x^n-x^{2n}|=|x^n(1-x^n)| \leq |x^n| \leq \epsilon$

Since $x$ cannot be eliminated, the convergence is not uniform.

Does this proof work?

  • 8
    $\begingroup$ Hint: $f_n(2^{-\frac{1}{n}}) = \frac14$ $\endgroup$ Apr 27, 2013 at 13:50
  • $\begingroup$ Here is a related problem. $\endgroup$ Apr 27, 2013 at 14:39
  • $\begingroup$ @achillehui please can you explain that; how you got $\frac14$ $\endgroup$ Jun 23, 2021 at 1:02

1 Answer 1


The sequence $(f_n)$ is pointwise convergent to the zero function $f$ on $[0,1]$ and by derivative we have $$f'_n(x)=x^{n-1}(n-2nx^n)=0\iff x=0\quad\text{or}\quad x=\frac{1}{2^n}$$ hence we find $$||f_n-f||_\infty=\sup_{x\in[0,1]}|f_n(x)-f(x)|=f_n\left(\frac{1}{2^n}\right)=\frac{1}{4}\not\to0$$ so $(f_n)$ does not converges uniformly to $f$ on $[0,1]$

Note We can prove that we have the uniform convergence of $(f_n)$ to $f$ on any interval of the form $[a,1]$ where $0<a<1$.


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