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As I was going over the definitions of pointwise and uniform convergence I came to the following problem: since the canonical example for continuous functions on $[0,1)$ which are pointwise but bot uniform convergent(wrt the constant function $f=0$) is sequence of functions $f_n(x) = x^n$ I ask myself is there such sequence of functions for the interval $[0,1]$. So far I couldn't think find any example and I am stating to believe that the answer might be negative. So what do think, is there such sequence and if not can it be proven that such sequence does not exist?

mechanodroid
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sve
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1 Answers1

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Certainly. take for example $f_n(x) = nxe^{-nx}$ which converges pointwise but not uniformly to $0$ on $[0,1]$. (And as pointed out in the comment, your own example works as well.)

mrf
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    Mine example does not work when $x=1$ but yours works brilliantly since for every $n$ we have that $f_n(\frac{1}{n}) = e^{-1}$ which breaks the uniform convergence requirement. How did you come up with this function? – sve May 13 '15 at 21:22
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    @lnwvr Start with any function $g(x)$ such that $g(0) = 0$ and $\lim_{x\to\infty} g(x) = 0$ and put $f_n(x) = g(nx)$. (Think graphically: how does the graph of $f_n$ compare to the graph of $g$?) If $g$ is not the zero function, you get pointwise but not uniform convergence to $0$ on any finite interval $[0,a]$. – mrf May 13 '15 at 21:24
  • wow! @ i speaketh out of my naivety. @sve, you have amazingly demonstrated the existence of an 'x=1/n' for each 'n' such that |f_n(x)-f(x)|=|(e^-1)-(0)|=e^-1, which will not work for any $\varepsilon < e^-1$. Super cool. – Krishan Nov 22 '20 at 07:40