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If a $\{f_n\}$ a sequence of continuous functions, converges pointwise to a discontinuous function $g$, does it imply that $\{f_n\}$ does not converges uniformly to $g$?

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  • $\begingroup$ f is a not a continuous function $\endgroup$ – user284275 Feb 27 '16 at 2:47
  • $\begingroup$ You mean a sequence of functions? $\endgroup$ – Qiaochu Yuan Feb 27 '16 at 2:54
  • $\begingroup$ oh yes, sorry if {fn} converges pointwise to a discontinuous function g, does it imply this sequence does not converge uniformly to g? $\endgroup$ – user284275 Feb 27 '16 at 2:57
  • $\begingroup$ It's easy for a sequence of discontinuous functions to converge uniformly to a discontinuous function. Perhaps you meant to assume that the functions $f_n$ are continuous functions? $\endgroup$ – bof Feb 27 '16 at 3:08
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    $\begingroup$ Use $\{f_n\}$ to show $\{f_n\}$. Formatting tips here. $\endgroup$ – Em. Feb 27 '16 at 3:32
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Suppose $\{f_n\}$ is a sequence of continuous functions such that $\{f_n\}$ converges pointwise to a discontinuous function $g$, then it must be the case that $\{f_n\}$ does not converge uniformly to $g$


Proof:

If $\{f_n\}$ is a uniformly convergent sequence of continuous functions, then by definition $\forall \epsilon > 0, \exists N$ s.t. $\forall x \in M, \forall n \geq N, |f_n(x) - f(x)| < \epsilon$

So if $f(x)$ is discontinuous at a point, then there exists some $x_o \in M$ and $\epsilon > 0$ such that $|f_n(x_o) - f(x_o)| \geq \epsilon$

Hence $\{f_n\}$ must not be uniformly convergent.


Example:

$f_n(x) = x^n$ on $[0,1]$

Each $f_n$ is continuous, and converges pointwise to $0, \forall x \in [0, 1)$ and $1, x =1$

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