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We know that $f_n:(0,1)\rightarrow\mathbb{R}$ is a sequence of nondecreasing functions on interval $(0,1)$ and that $f:(0,1)\rightarrow \mathbb{R}$ is a continuous function. Let $A\subset (0,1)$ be a dense subset of $(0,1)$, which fulfills the condition $$ \forall_{x\in A} \ \lim_{n\rightarrow \infty}f_n(x)=f(x).$$ With this assumptions I have to

a) show that $f_n$ converges uniformly to $f$ on every $[a,b]\subset(0,1)$,

b) answer if $f_n$ converges uniformly to $f$ on $(0,1)$.

My attempts:

a) We have to show that $$ \forall_{\epsilon>0} \ \exists_{n_0=n_0(\epsilon)} \forall_{x\in [a,b]}\forall_{n>n_0} \ \ |f(x)-f_n(x)|<\epsilon.$$ For $f_n(x)$, $x\in A$, where $A$ is a dense subset of $(0,1)$ we can observe pointwise convergence, so let's take $[a,b]$ in such a way to make it a dense subset of $(0,1)$. Then for every $\epsilon$ we will be able to choose such $n_0=n_0(\epsilon)$ that $\forall_{x\in [a,b]}\forall_{n>n_0} \ \ |f(x)-f_n(x)|<\epsilon$ (from the point convergence).

b) $f_n$ doesn't converge uniformly to $f$ on $(0,1)$ because on this interval doesn't converge even pointwise.

Are my solutions sensible? If not, I politely ask for help. Thanks in advance and have a good day.

Novice
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1 Answers1

2

a)

  • First, we show pointwise convergence. For any $x$ we build increasing $x_k \to x$ and decreasing $x'_k \to x$, belonging to $A$. For any $n$, we have $$f_n(x_k) \le f_n(x) \le f_n(x'_k)$$ Since $f_n$ converges on $A$, $$\forall k\ \forall \epsilon\ \exists n_{k,\epsilon}:\ \forall n > n_{k,\epsilon} \ \ |f_n(x_k) - f(x_k)| < \frac \epsilon 2 \tag{1}\label{eq1}$$ On the other hand, since $f$ is continuous: $$\forall \epsilon\ \exists k_{\epsilon}:\ \forall k \ge k_{\epsilon} \ \ |f(x_k) - f(x)| < \frac \epsilon 2 \tag{2}\label{eq2}$$ (The same for $x'_k$). Therefore, $\forall n > n_{k_\epsilon,\epsilon}$ \begin{alignat}{1} \text{(by \eqref{eq1} for $k \gets k_\epsilon$)} \quad\quad & f(x_{k_\epsilon}) &- \frac \epsilon 2 &\le f_n(x) &\le f(x'_{k_\epsilon}) &+ \frac \epsilon 2 \Rightarrow \\ \text{(by \eqref{eq2} for $k \gets k_\epsilon$)} \quad\quad & f(x) &- \epsilon &\le f_n(x) &\le f(x) &+ \epsilon \end{alignat} I.e., we have a pointwise convergence at point $x$.

  • Uniform convergence on $[a,b]$ is shown here.

b) It's false: consider $f_n(x) = \begin{cases} -\frac 1x, & x < \frac 1n \\ 0, & x \ge \frac 1n \end{cases}$. For each $x$, we have $f_n(x) \to f(x) = 0$. But convergence is not uniform: for any $n$ and $\epsilon$ there exists $x = \min(\frac 1 {2n}, \frac 1 {2 \epsilon})$: $f_n(x) \le -2 \epsilon$.

Dmitry
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  • Thanks. Can I ask from which property have we the last inequality in b)? If I am guessing correctly it is something like $f_n(x)\leq f(x)$. – Novice Aug 02 '20 at 08:07
  • Well, I just substituted $x$ into $f_n(x)$... I guess I’ve used that $f_n$ is monotone – Dmitry Aug 02 '20 at 12:20
  • @Dmitry If $k\gt k_{\epsilon}$ then does $|f_n(x_k)-f (x_k)|\lt \epsilon/2 $ hold for $ n\gt n_{k_{\epsilon},\epsilon}?$ because I think you are using the above (unclaimed) fact to prove the second last inequality in proving pointwise convergence at point $x$. If I am wrong , then please justify how do you get the inequality I mentioned. – user-492177 Aug 05 '20 at 11:29
  • @Dmitry According to me, your inequality is still valid by taking $n\gt n_{k_{\epsilon},\epsilon}$ and writing $f_n(x) \ge f_n(x_{k_{\epsilon}}) \ge f(x_{k_{\epsilon}})-\frac{\epsilon}2 \ge f(x)-\epsilon$. Is this wrong ? – user-492177 Aug 05 '20 at 11:36
  • @user710290, thanks! I've fixed it. it suffices to just consider $k_\epsilon$ itself. – Dmitry Aug 05 '20 at 18:20
  • @Dmitry Thanks for the clarification. +1 from me! – user-492177 Aug 05 '20 at 18:36