2
$\begingroup$

Briefly, the definitions of point-wise convergence (PWC) and uniform convergence (UC) for a sequence of functions $f_n:[a,b]\to\mathbb{R}$ in my mind are recorded as

\begin{align*} &\text{Point Wise Convergent on $[a,b]$} \iff \\ &\forall x\in [a,b]\,\forall\epsilon\gt0\,\exists N=\mathcal{N}(\epsilon,x)\gt0,\,n\ge N \implies |f_n(x)-f(x)|<\epsilon \\ \\ &\text{Uniformly Convergent on $[a,b]$}\iff \\ &\forall x\in [a,b]\,\forall\epsilon\gt0\,\exists N=\mathcal{N}(\epsilon)\gt0,\quad\, n\ge N \implies |f_n(x)-f(x)|<\epsilon. \end{align*}

So the difference is that in PWC the number $N$ depends on $x$ while in UC it does not which means just one $N$ works for all $x$ in $[a,b]$.

I want to prove the following theorem.

Theorem. If the functions $f_n:[a,b]\to\mathbb{R}$ are continuous at $x_0\in[a,b]$ and their sequence converges uniformly to the function $f:[a,b]\to\mathbb{R}$ on $[a,b]$ then $f$ is continuous at $x_0$.

Proof. According to the definition of continuity at $x_0$ for $f$, we want to show that

\begin{align*} \forall\epsilon\gt0\,\exists \delta=\Delta(\epsilon,x_0)\gt0,\,|x-x_0|<\delta \implies |f(x)-f(x_0)|<\epsilon. \end{align*}

According to triangle inequality we have

\begin{align*} |f(x)-f(x_0)|\le|f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|. \tag{1} \end{align*}

If we could control each of the three terms on the RHS of $(1)$ such that they were less than $\frac{\epsilon}{3}$ then the theorem was proved. According to the assumptions we know that the following holds

\begin{align*} &\forall\epsilon_1\gt0\,\exists \delta_1=\Delta_1(\epsilon_1,x_0,n)\gt0,\,|x-x_0|<\delta_1 \implies |f_n(x)-f_n(x_0)|<\epsilon_1 \\ \\ &\forall x\in [a,b]\,\forall\epsilon_2\gt0\,\exists N=\mathcal{N}(\epsilon_2)\gt0, n\ge N \implies |f_n(x)-f(x)|<\epsilon_2. \end{align*} Finally, choosing any $\epsilon_1$ and $\epsilon_2$ such that $0<\epsilon_1\le\frac{\epsilon}{3}$ and $0<\epsilon_2\le\frac{\epsilon}{3}$ and setting any $\delta$ such that $\delta\le\delta_1$ will do the job. For simplicity, one can usually take the equality cases which means $\epsilon_1=\epsilon_2=\frac{\epsilon}{3}$ and $\delta=\delta_1$.

$1$. Is my proof OK? Any suggestions for improvement is really appreciated.

$2$. Are the notations $\mathcal{N}(\epsilon,x)$ or $\Delta(\epsilon,x,n)$ OK? I just employed them to emphasize the the dependence on $\epsilon$ and $x$. Any better suggestion is welcomed.

$3$. I was wondering which step would fail if we just had PWC? An example can be helpful.

$\endgroup$
1
$\begingroup$

Your notations and proof seem great, and why the condition PWC is not sufficient is that under this you cannot choose your $\mathcal{N}(\epsilon_2)$ feasible for any $x$ in your domain. (Maybe for arbitrarily large $N$ there always exist some $x$ near $x_0$ making your argument fail.)

$\endgroup$
  • $\begingroup$ Thanks. :) Is there any weaker condition than uniform convergence on $[a,b]$. For example, can't we say that $f_n$ be uniformly convergent in some neighborhood of $x_0$? I don't know maybe some notion like locally uniform convergent or something may still work. $\endgroup$ – H. R. Sep 12 '17 at 11:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.