Show that the limit of f is equal to y

[evinda]

Hello! ;)
Suppose that $A \subseteq \mathbb{R}, f_n \to f$ uniformly in $A$, $x_0$ an accumulation point of $A$ and $ \displaystyle \lim_{x \to x_0} {f_n(x)}=y_n \in \mathbb{R}$. $(y_n)$ converges in $\mathbb{R}$.Show that $ \displaystyle \lim_{x \to x_0} {f(x)}= \lim_{n \to +\infty} {y_n} \Rightarrow \lim_{x \to x_0} \lim_{n \to +\infty} {f_n(x)}= \lim_{n \to +\infty} \lim_{x \to x_0} f_n(x)$.

That is the solution that the assistant of the professor gave us:

• We set $ \displaystyle y= \lim_{n \to + \infty} y_n$
  
  We have $|f(x)-y|=|f(x)-y_n+y_n-y| \leq |f(x)-y_n|+|y_n-y|$
  
  $ \displaystyle \Rightarrow \lim_{x \to x_0} \sup_{x}{|f(x)-y|} \leq \lim_{x \to x_0} \sup_{x} |f(x)-y_n|+ \lim_{x \to x_0} \sup_{x} |y_n-y| \overset{n \to \infty}{\longrightarrow} 0+0=0$
  
  We have used that $ \displaystyle \lim_{x \to x_0} \sup_{x} {|f(x)-y_n|}=0$,because:
  
  $|f(x)-y_n|=|f(x)-f_n(x)+f_n(x)-y_n| \leq |f(x)-f_n(x)|+|f_n(x)-y_n|$
  
  $ \displaystyle \lim_{ x \to x_0} \sup_{x} |f(x)-y_n| \leq \lim_{x \to x_0} \sup_{x} |f(x)-f_n(x)|+ \lim_{x \to x_0} \sup_{x} |f_n(x)-y_n|$.
  
  But, $ \displaystyle \lim_{x \to x_0} \sup_{x} f(x)= \sup_{x} \{ \lim_{n \to + \infty}{f(z_n)} | z_n \to x_0 \}$
  
  Therefore, we have $ \displaystyle \lim_{x \to x_0} \sup_{x} |f(x)-y_n| \leq \sup_{x} |f(x)-f_n(x)|\overset{n \to \infty}{\longrightarrow} 0 $Is this right? I am not really sure... I would do it like that:
• We set $ \displaystyle y= \lim_{n \to + \infty} y_n$
  
  We want to show : $ \displaystyle \lim_{x \to x_0} f(x)= \lim_{n \to +\infty} y_n \Rightarrow \lim_{x \to x_0}f(x)=y$
  
  $|f(x)-y|=|f(x)-y_n+y_n-y| \leq |f(x)-y_n|+|y_n-y| (1)$
  
  Let $\epsilon>0$. As $y_n \to y, \exists n_0$ such that $ \forall n \geq n_0: |y_n-y|< \epsilon$
  
  $ \displaystyle |f(x)-y_n|=|f(x)-f_n(x)+f_n(x)-y_n| \leq |f_n(x)-f(x)|+|f_n(x)-y_n|$
  
  $ \displaystyle \leq \sup_{x} |f_n(x)-f(x)| + |f_n(x)-y_n| \leq \epsilon $, because:
  
  $f_n \to f$ uniformly in $A$, so $ \displaystyle \sup_{x} |f_n(x)-f(x)| \leq \epsilon$.
  Also, $ \displaystyle \lim_{x \to x_0} {f_n(x)}=y_n \in \mathbb{R}$, so $|f_n(x)-y_n|< \epsilon$.
  
  From the relation $(1)$,we have $|f(x)-y| \leq \epsilon$.

Could you tell me if it is right?

 

[Opalg]

Hello! ;)
Suppose that $A \subseteq \mathbb{R}, f_n \to f$ uniformly in $A$, $x_0$ an accumulation point of $A$ and $ \displaystyle \lim_{x \to x_0} {f_n(x)}=y_n \in \mathbb{R}$. $(y_n)$ converges in $\mathbb{R}$.Show that $ \displaystyle \lim_{x \to x_0} {f(x)}= \lim_{n \to +\infty} {y_n} \Rightarrow \lim_{x \to x_0} \lim_{n \to +\infty} {f_n(x)}= \lim_{n \to +\infty} \lim_{x \to x_0} f_n(x)$.

That is the solution that the assistant of the professor gave us:

• We set $ \displaystyle y= \lim_{n \to + \infty} y_n$
  
  We have $|f(x)-y|=|f(x)-y_n+y_n-y| \leq |f(x)-y_n|+|y_n-y|$
  
  $ \displaystyle \Rightarrow \lim_{x \to x_0} \sup_{x}{|f(x)-y|} \leq \lim_{x \to x_0} \sup_{x} |f(x)-y_n|+ \lim_{x \to x_0} \sup_{x} |y_n-y| \overset{n \to \infty}{\longrightarrow} 0+0=0$
  
  We have used that $ \displaystyle \lim_{x \to x_0} \sup_{x} {|f(x)-y_n|}=0$,because:
  
  $|f(x)-y_n|=|f(x)-f_n(x)+f_n(x)-y_n| \leq |f(x)-f_n(x)|+|f_n(x)-y_n|$
  
  $ \displaystyle \lim_{ x \to x_0} \sup_{x} |f(x)-y_n| \leq \lim_{x \to x_0} \sup_{x} |f(x)-f_n(x)|+ \lim_{x \to x_0} \sup_{x} |f_n(x)-y_n|$.
  
  But, $\color{red}{ \displaystyle \lim_{x \to x_0} \sup_{x} f(x)= \sup_{x} \{ \lim_{n \to + \infty}{f(z_n)} | z_n \to x_0 \}}$
  
  Therefore, we have $ \displaystyle \lim_{x \to x_0} \sup_{x} |f(x)-y_n| \leq \sup_{x} |f(x)-f_n(x)|\overset{n \to \infty}{\longrightarrow} 0 $Is this right? I am not really sure... I would do it like that:
• We set $ \displaystyle y= \lim_{n \to + \infty} y_n$
  
  We want to show : $ \displaystyle \lim_{x \to x_0} f(x)= \lim_{n \to +\infty} y_n \Rightarrow \lim_{x \to x_0}f(x)=y$
  
  $|f(x)-y|=|f(x)-y_n+y_n-y| \leq |f(x)-y_n|+|y_n-y| (1)$
  
  Let $\epsilon>0$. As $y_n \to y, \exists n_0$ such that $ \forall n \geq n_0: |y_n-y|< \epsilon$
  
  $ \displaystyle |f(x)-y_n|=|f(x)-f_n(x)+f_n(x)-y_n| \leq |f_n(x)-f(x)|+|f_n(x)-y_n|$
  
  $ \displaystyle \leq \sup_{x} |f_n(x)-f(x)| + |f_n(x)-y_n| \leq \epsilon $, because:
  
  $f_n \to f$ uniformly in $A$, so $ \displaystyle \sup_{x} |f_n(x)-f(x)| \leq \epsilon$.
  Also, $ \displaystyle \lim_{x \to x_0} {f_n(x)}=y_n \in \mathbb{R}$, so $|f_n(x)-y_n|< \epsilon$.
  
  From the relation $(1)$,we have $|f(x)-y| \leq \epsilon$.

Could you tell me if it is right?

Your proof looks fine (except that the very last $ \epsilon$ looks as though it should be $2 \epsilon$). The other proof goes along the same lines apart from the equation that I have coloured red. I do not see what is going on there, and I don't trust it. In any case, it does not make clear where the uniform convergence $f_n\to f$ is used, unlike your solution which pinpoints that essential component of the proof.

 

[evinda]

Your proof looks fine (except that the very last $ \epsilon$ looks as though it should be $2 \epsilon$). The other proof goes along the same lines apart from the equation that I have coloured red. I do not see what is going on there, and I don't trust it. In any case, it does not make clear where the uniform convergence $f_n\to f$ is used, unlike your solution which pinpoints that essential component of the proof.

Great!Thank you very much! :)