Math Genius: Dini’s theorem (specific case)

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Note: I asked this question before but it wasn’t well written, So I deleted my previous question and re-wrote it.

According to Dini’s theorem:

If $X$ is a compact topological space, and ${ f_n }$ is a monotonically
increasing sequence (meaning $f_n(x) leq f_{n+1}(x)$ for all $n$ and $x$) of
continuous real-valued functions on $X$ which converges pointwise to a
continuous function $f$, then the convergence is uniform.

The same conclusion holds if ${ f_n }$ is monotonically decreasing
instead of increasing.

(Note: I have proven both cases)

But, what if for every $n$ ${f_n(x0)}$ is monotonic but for some values of $n$ it’s monotonically decreasing and for other it’s monotonically decreasing.
for example; for all even values it is increasing and for non-even values it is decreasing.

How could I prove that Dini’s theorem is effective in this case?
In other words, how to prove that the convergence is uniform

Let $A={x: f_n(x) leq f_{n+1}(x) forall n}$ and $B={x: f_n(x) geq f_{n+1}(x) forall n}$. Note that $A$ and $B$ are closed sets and hence they are also compact. Also $A cup B=X$. $f_n to f$ uniformly on each of these sets. Given $epsilon >0$ there exist $n_1, n_2$ such that $|f_n(x)-f(x)| <epsilon$ for all $x in A$ for all $n >n_1$ and $|f_n(x)-f(x)| <epsilon$ for all $x in B$ for all $n >n_1$. Let $n_0=max {n_1,n_2}$. Then $|f_n(x)-f(x)| <epsilon$ for all $x in X$ for all $n >n_0$.

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Math Genius: Uniform converge, using the Supremum theorem (want to make sure I solved correctly)

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I have the givens: $$[a,b], [c,d] subseteq mathbb{R}quad f_n:[a,b]longrightarrow [c,d] quad g:[c,d] longrightarrow mathbb{R}$$ $g$ is continuous and $f_n$ is uniformly converge on $[a,b]$ to $f$. I need to prove that $h_n=g(f_n(x)) is$ is uniformly converge to $h=g(f(x))$ on $[a,b]$

I showed that because $g$ continuous so according to heine deffinition $lim_{n to infty} g(f_n(x))=g(f(x))$.
from here I wrote $lim_{n to infty}sup|g(f_n(x))-g(f(x))|=0.$

The thing is I am not sure if I can argue the last equation, because I “bring” $n$ to infinity after I found the supremum.

Just because, for each individual $x$, you have $lim_{ntoinfty}bigl|gbigl(f_n(x)bigr)-gbigl(f(x)bigr)bigr|=0$, you cannot jump to $$lim_{ntoinfty}sup_{xin[a,b]}bigl|gbigl(f_n(x)bigr)-gbigl(f(x)bigr)bigr|=0.$$

Note that, since the domain of $g$ is an interval which is closed and bounded, $g$ is uniformly continuous. Now, take $varepsilon>0$. There is a $delta>0$ such that$$|x-y|<deltaimpliesbigl|g(x)-g(y)bigr|<varepsilon.$$And there is a $NinBbb N$ such that$$ngeqslant Nimplies(forall xin[a,b]):bigl|f_n(x)-f(x)bigr|<delta,$$since $(f_n)_{ninBbb N}$ converges uniformly to $f$. So, if $ngeqslant N$ and if $xin[a,b]$, then$$bigl|gbigl(f_n(x)bigr)-gbigl(f(x)bigr)bigr|<varepsilon.$$

hint

Use Heine’s Theorem, which states that

$g$ beeing continuous at the compact $ [c,d] $, is Uniformly continuous at $ [c,d]$.

Then, you can use Cauchy’s criteria for the uniform convergence.

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Math Genius: $f_n$ uniform converge to f in $[a,b)$ means it uniform converge in $[a,b]$?

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I have the given : $f_n$ uniform converge to $f$ in $[a,b)$ I am asked for the following:

  1. if $f_n$ is Continuous for every $ninmathbb{N}$ is it true that $f_n$ uniform converge to $f$ in $[a,b]$?

  2. if both $f$ and $f_n$ are Continuous is it true that $f_n$ uniform converge to $f$ in ?

I not sure I understand how to go from $[a,b)$ to $[a,b]$ thats is the main trick I think required here here

Suppose that $f_ncolon[a,b]longrightarrowBbb R$ is defined as $f_n(x)=0$ and that you have$$begin{array}{rccc}fcolon&[a,b]&longrightarrow&Bbb R\&x&mapsto&begin{cases}0&text{ if }x<b\1&text{ otherwise.}end{cases}end{array}$$Then $(f_n)_{ninBbb N}$ converges uniformly to $f$ on $[a,b)$, but not on $[a,b]$.

However, if $f$ is continuous then, yes, uniform convergence on $[a,b)$ implies uniform convergence on $[a,b]$. Take $varepsilon>0$ and take $NinBbb N$ such that, if $ngeqslant N$ and if $xin[a,b)$, then $|f_n(x)-f(x)|<fracvarepsilon2$. Then, by the continuity of all the functions involved, if $ngeqslant N$ and if $xin[a,b]$, then $|f_n(x)-f(x)|leqslantfracvarepsilon2<varepsilon$.

This is true. In fact, if $F: [a;b] to Bbb R$ is any continuous function, then $$sup_{x in [a;b]} |F(x)| = sup_{x in [a;b)} |F(x)|$$ Now, since uniform convergence is defined using the supremum, there is no distinction between $[a,b]$ and $[a,b)$. In other words
$$lim_n sup_{x in [a;b]} |f_n(x)-f(x)| = lim_n sup_{x in [a;b)} |f_n(x)-f(x)|$$

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Math Genius: prove that $lim_{ntoinfty} int_a^b |f(x) – f_n(x)|^p dx =0$ for every $p>0$

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Let $(f_n)$ be a sequence of continues functions in $[a, b]$ that uniformly converge to $f(x)$
I need to prove that $$lim_{ntoinfty} int_a^b |f(x) – f_n(x)|^p dx =0$$ for every $p>0$.


Let $g_n(x)=|f(x) – f_n(x)|^p$, because $(f_n)$ are continues functions in $[a, b]$ so does $f(x)$ and as a result $g_n(x)$ too. thus $$lim_{ntoinfty}g_n(x)=lim_{ntoinfty}|f(x) – f_n(x)|^p=0=g(x)$$ In order to prove that $lim_{ntoinfty} int_a^b |f(x) – f_n(x)|^p dx =0$ for every $p>0$, I first need to prove that $g_n(x)$ uniformly converge to $g(x)$. $(f_n)$ uniformly converge to $f(x)$ hence $$lim_{ntoinfty} sup{|f_n(x)-f(x)|, ale xle b}=0$$ so there is an $N$ that for every $n>N$ $$|f_n(x)-f(x)|le sup{|f_n(x)-f(x)|, ale xle b}<1$$ let’s assume $pge 1$ for every $n>N$ $$|f_n(x)-f(x)|^ple |f_n(x)-f(x)|Longrightarrow sup{|f_n(x)-f(x)|^p, ale xle b} le sup{|f_n(x)-f(x)|, ale xle b}$$ hence $$lim_{ntoinfty} sup{|f_n(x)-f(x)|^p, ale xle b}=0$$ and $g_n(x)$ uniformly converge to $g(x)$. but this is not the case for $0<p<1$, and I haven’t found a way to show it does uniformly converge for said p.

Let $pin(0,infty)$. Fix $varepsilon>0$. There exists a $Ninmathbb N$ such that $sup_{[a,b]}|f(x)-f_n(x)|<varepsilon^{1/p}$ for all $n>N$. Thus for any $xin [a,b]$ and $n>N$ we have $|f(x)-f_n(x)|^p<varepsilon$. As this is true for any $x$ independent of $N$ we have $|f(x)-f_n(x)|^pto 0$ uniformly.

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Math Genius: prove that $lim_{ntoinfty} int_a^b |f(x) – f_n(x)|^p dx =0$ for every $p>0$

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Let $(f_n)$ be a sequence of continues functions in $[a, b]$ that uniformly converge to $f(x)$
I need to prove that $$lim_{ntoinfty} int_a^b |f(x) – f_n(x)|^p dx =0$$ for every $p>0$.


Let $g_n(x)=|f(x) – f_n(x)|^p$, because $(f_n)$ are continues functions in $[a, b]$ so does $f(x)$ and as a result $g_n(x)$ too. thus $$lim_{ntoinfty}g_n(x)=lim_{ntoinfty}|f(x) – f_n(x)|^p=0=g(x)$$ In order to prove that $lim_{ntoinfty} int_a^b |f(x) – f_n(x)|^p dx =0$ for every $p>0$, I first need to prove that $g_n(x)$ uniformly converge to $g(x)$. $(f_n)$ uniformly converge to $f(x)$ hence $$lim_{ntoinfty} sup{|f_n(x)-f(x)|, ale xle b}=0$$ so there is an $N$ that for every $n>N$ $$|f_n(x)-f(x)|le sup{|f_n(x)-f(x)|, ale xle b}<1$$ let’s assume $pge 1$ for every $n>N$ $$|f_n(x)-f(x)|^ple |f_n(x)-f(x)|Longrightarrow sup{|f_n(x)-f(x)|^p, ale xle b} le sup{|f_n(x)-f(x)|, ale xle b}$$ hence $$lim_{ntoinfty} sup{|f_n(x)-f(x)|^p, ale xle b}=0$$ and $g_n(x)$ uniformly converge to $g(x)$. but this is not the case for $0<p<1$, and I haven’t found a way to show it does uniformly converge for said p.

Let $pin(0,infty)$. Fix $varepsilon>0$. There exists a $Ninmathbb N$ such that $sup_{[a,b]}|f(x)-f_n(x)|<varepsilon^{1/p}$ for all $n>N$. Thus for any $xin [a,b]$ and $n>N$ we have $|f(x)-f_n(x)|^p<varepsilon$. As this is true for any $x$ independent of $N$ we have $|f(x)-f_n(x)|^pto 0$ uniformly.

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Math Genius: Termwise differentiation of series

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THEOREM 17.5 .2 Termwise Differentiation of Series
Let $sum_{n=1}^{infty} a_{n}(x)$ converge on an $x$ interval $I$. Then
$$
frac{d}{d x} sum_{n=1}^{infty} a_{n}(x)=sum_{n=1}^{infty} frac{d}{d x} a_{n}(x)
$$

if the series on the right converges uniformly on $I$. (The theorem is from the book “Engineering Mathematics, M.D. Greenberg 2nd edition, page 875.)

My question is why uniform convergence allows us to have such equality?
Initially I thought distributivity allows such equality needs to hold but I couldn’t understand why uniform convergence have such superiority. Also, could you give an example of such series that holds the given equality but not uniformly convergent. Thanks in advance.

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Math Genius: $frac{n}{2}int_{x-1/n}^{x+1/n} f(t)dt$ converges uniformly to $f$ where $f:Bbb Rto Bbb R$ is continuous

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Suppose $f$ is a continuous function $Bbb Rto Bbb R$, and for each $nin Bbb N$ define $f_n:Bbb Rto Bbb R$ by $f_n(x)=frac{n}{2}int_{x-1/n}^{x+1/n} f(t)dt$. It is clear that $f_n$ converges to $f$ pointwise, by the continuity of $f$. But does $f_n$ converges to $f$ also uniformly? The answer would be positive if $f$ is uniformly continuous, but I can’t see whether the answer is still positive in the general case.

You can explicitly compute the value of this expression when $f(t)=e^{t}$ and see that the convergence is not uniform on $mathbb R$.

Note that $e^{x} a_n$ can never tend to $0$ uniformly on $mathbb R$ for sequence of positive numbers $a_n$. In this example $f_n(x)-f(x)$ has this form.

PS: I just noticed that $f(t)=t^{k}$ is also a counter-example for every $k >2$.

A correct counterexample is $f(x)=e^x$. Note $$e^x – n/2(exp(x+1/n)-exp(x-1/n))=e^x(1-n/2(exp(1/n) – exp(-1/n)))$$

and $1 – n/2(exp(1/n) – exp(-1/n)) to 0$ as $n to infty$. Since $e^x to infty$ as $x to infty$, the $N$ required to force $e^x(1-n/2(exp(1/n) – exp(-1/n)))<epsilon$ for $n geq N$ very much depends on $x$.

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Math Genius: $frac{n}{2}int_{x-1/n}^{x+1/n} f(t)dt$ converges uniformly to $f$ where $f:Bbb Rto Bbb R$ is continuous

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Suppose $f$ is a continuous function $Bbb Rto Bbb R$, and for each $nin Bbb N$ define $f_n:Bbb Rto Bbb R$ by $f_n(x)=frac{n}{2}int_{x-1/n}^{x+1/n} f(t)dt$. It is clear that $f_n$ converges to $f$ pointwise, by the continuity of $f$. But does $f_n$ converges to $f$ also uniformly? The answer would be positive if $f$ is uniformly continuous, but I can’t see whether the answer is still positive in the general case.

You can explicitly compute the value of this expression when $f(t)=e^{t}$ and see that the convergence is not uniform on $mathbb R$.

Note that $e^{x} a_n$ can never tend to $0$ uniformly on $mathbb R$ for sequence of positive numbers $a_n$. In this example $f_n(x)-f(x)$ has this form.

PS: I just noticed that $f(t)=t^{k}$ is also a counter-example for every $k >2$.

A correct counterexample is $f(x)=e^x$. Note $$e^x – n/2(exp(x+1/n)-exp(x-1/n))=e^x(1-n/2(exp(1/n) – exp(-1/n)))$$

and $1 – n/2(exp(1/n) – exp(-1/n)) to 0$ as $n to infty$. Since $e^x to infty$ as $x to infty$, the $N$ required to force $e^x(1-n/2(exp(1/n) – exp(-1/n)))<epsilon$ for $n geq N$ very much depends on $x$.

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Math Genius: How do we prove that a power series centered at $aintextbf{R}$ is continuous on the interval $(a-R,a+R)$?

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Let $f:(a-R,a+R)totextbf{R}$ be the function
begin{align*}
f(x) = sum_{n=0}^{infty}c_{n}(x-a)^{n}
end{align*}

For any $0 < r < R$, the series $displaystylesum_{n=0}^{infty}c_{n}(x-a)^{n}$ converges uniformly to $f$ on the compact interval $E = [a-r,a+r]$.

In particular, $f$ is continuous on $(a-R,a+R)$.

My solution

Let us define $f_{n}(x) = c_{n}(x-a)^{n}$ on $E$. Then each $f_{n}$ is a continuous and bounded real-valued function. Hence we have that
begin{align*}
|f_{n}| = sup_{xin E}|f_{n}(x)| = |c_{n}|r^{n} Rightarrow sum_{n=0}^{infty}|f_{n}| = sum_{n=0}^{infty}|c_{n}|r^{n} = sum_{n=0}^{infty}|c_{n}r^{n}|
end{align*}

Since $0 < r < R$, the last numerical series converges.

Consequently, due to the Weierstrass $M$-test, the proposed power series converges uniformly on $E$, whence $f$ is continuous.

My question is: how do I prove from the obtained results that $f$ is continuous on the open interval $(a-R,a+R)$?

Take $xin(a-R,a+R)$. Then $|x-a|<R$. Take some $rinbigl(|x-a|,Rbigr)$. Then $f$ is continuous in $[a-r,a+r]$ and, in particular, in $(a-r,a+r)$, to which $x$ belongs. So, $f$ is continuous at $x$.

hint

Take $x_0$ in $(a-R,a+R)$.

Assume that $a<x_0<a+R$.

Put $$y=x_0+frac{a+R-x_0}{2}$$
then

$$x_0<y<a+R$$

$f$ is continuous at $[-y,y]$, so it is continuous at $x_0$.

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Math Genius: How do we prove that a power series centered at $aintextbf{R}$ is continuous on the interval $(a-R,a+R)$?

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Let $f:(a-R,a+R)totextbf{R}$ be the function
begin{align*}
f(x) = sum_{n=0}^{infty}c_{n}(x-a)^{n}
end{align*}

For any $0 < r < R$, the series $displaystylesum_{n=0}^{infty}c_{n}(x-a)^{n}$ converges uniformly to $f$ on the compact interval $E = [a-r,a+r]$.

In particular, $f$ is continuous on $(a-R,a+R)$.

My solution

Let us define $f_{n}(x) = c_{n}(x-a)^{n}$ on $E$. Then each $f_{n}$ is a continuous and bounded real-valued function. Hence we have that
begin{align*}
|f_{n}| = sup_{xin E}|f_{n}(x)| = |c_{n}|r^{n} Rightarrow sum_{n=0}^{infty}|f_{n}| = sum_{n=0}^{infty}|c_{n}|r^{n} = sum_{n=0}^{infty}|c_{n}r^{n}|
end{align*}

Since $0 < r < R$, the last numerical series converges.

Consequently, due to the Weierstrass $M$-test, the proposed power series converges uniformly on $E$, whence $f$ is continuous.

My question is: how do I prove from the obtained results that $f$ is continuous on the open interval $(a-R,a+R)$?

Take $xin(a-R,a+R)$. Then $|x-a|<R$. Take some $rinbigl(|x-a|,Rbigr)$. Then $f$ is continuous in $[a-r,a+r]$ and, in particular, in $(a-r,a+r)$, to which $x$ belongs. So, $f$ is continuous at $x$.

hint

Take $x_0$ in $(a-R,a+R)$.

Assume that $a<x_0<a+R$.

Put $$y=x_0+frac{a+R-x_0}{2}$$
then

$$x_0<y<a+R$$

$f$ is continuous at $[-y,y]$, so it is continuous at $x_0$.

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