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Let $f_n, f : D \to \mathbb{R}$ with $A(\neq \phi) \subset \mathbb{R}$

Here the $f_n$ are the sequence of the continuous functions.

I already knew the fact "$f_n$ is uniformly converge to $ f$ on $D$ $\Rightarrow$ $f$ is continuous on $D$"

But There is a lecturer's claim that

Say $f_n$ is continuous on $A$

"$f_n$ is uniformly converge to $ f$ on $A$ $\not\Rightarrow$ $f$ is continuous on $A$" (statement (*))

He suggested the counter-example as like the below.

$f_n, f : \mathbb{R} \to \mathbb{R}$ with $A(\neq \phi) \subset \mathbb{R}$

Here the $D = \mathbb{R}, A = \mathbb{Q}^c$ and

Since the rational number set,$\mathbb{Q}$ is countable so $\mathbb{Q} = \{x_1, x_2, .... \}$

Then Defining each functions like the below

$f_n(x) = \begin{cases}1 & \text{$x \in \{ x_1, x_2, ...,x_n\}$} \\ 0 & \text{$o.w.$}\end{cases}$

$f(x) = \begin{cases} 1 & \text{$x \in \mathbb{Q} $ } \\ 0 & \text{$x \in \mathbb{Q}^c$} \end{cases}$

From here, my question begins.

First question) I put in to the sequence $x_{n+1} \in \mathbb{Q}$ to $\Vert f_n(x) - f(x) \Vert$

Then $\Vert f_n(x_{n+1}) - f(x_{n+1}) \Vert$ = $\Vert 0-1 \Vert = 1 \geq \epsilon(= {1 \over 2}) $

So my conclusion is this counterexample is totally wrong, since $f_n$ is not uniformly converge to $f$. What do you think about that? Is my thought wrong?

Second question) Like the lecture's thought, I guess the statement (*) is false. But I can't find any counterexample. If the lecture's counterexample is false, Please give me a counterexample.

p.s.) If the statement is true, Why does the statement(*) hold?

Thanks.

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There seems to be no connection between $D$ and $A$ in $(*)$. So $(*)$ is correct. For one example take $D=[0, \frac 12 ], A=[0,1]$, $f_n(x)=x^{n}, f(x)=0$ for $ x<1$, $f(1)=1$. Then $f_n \to f$ uniformly on $D$ and $f$ is not continuous on $A$.

However, the example given by the lecturer is wrong since $f_n$ does not converge uniformly to $f$ on $D=\mathbb R$.

Answer for the revised question: If $f_n \to f$ uniformly on $A$ and each $f_n$ is continuous on $A$ then the restriction of $f$ to $A$ is continuous. But you cannot say that $f$ is continuous at points of $A$ as in the example by your lecturer.

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  • $\begingroup$ Thanks for reply, Dear Mr,@Murthy. But I should have been asked "$f_n$ is uniformly converge to $ f$ on $A$" in statement (*) not "$f_n$ is uniformly converge to $ f$ on $D$" Sorry for typo. I edited my question. $\endgroup$ – se-hyuck yang Jan 28 '20 at 8:25
  • $\begingroup$ @se-hyuckyang I have edited my answer. $\endgroup$ – Kavi Rama Murthy Jan 28 '20 at 8:25
  • $\begingroup$ Ah yes! Thank you. :) $\endgroup$ – se-hyuck yang Jan 28 '20 at 8:27
  • $\begingroup$ What about the case if we add the condition $A \subset D$? Though adding more condition, his claim and counterexample are still not hold, right? $\endgroup$ – se-hyuck yang Jan 28 '20 at 14:31
  • $\begingroup$ @se-hyuckyang Actually $f_n(x)=f(x)=0$ for all $x$ in $A$ in his example. So $f_n \to f$ uniformly on $A$. $\endgroup$ – Kavi Rama Murthy Jan 28 '20 at 23:15

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