Math Genius: Dini’s theorem (specific case)

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

(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 $$|f_n(x)-f(x)| for all $$x in A$$ for all $$n >n_1$$ and $$|f_n(x)-f(x)| for all $$x in B$$ for all $$n >n_1$$. Let $$n_0=max {n_1,n_2}$$. Then $$|f_n(x)-f(x)| for all $$x in X$$ for all $$n >n_0$$.

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

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$$|x-y|And there is a $$NinBbb N$$ such that$$ngeqslant Nimplies(forall xin[a,b]):bigl|f_n(x)-f(x)bigr|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|

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]\$?

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$$begin{array}{rccc}fcolon&[a,b]&longrightarrow&Bbb R\&x&mapsto&begin{cases}0&text{ if }xThen $$(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 $$|f_n(x)-f(x)|. Then, by the continuity of all the functions involved, if $$ngeqslant N$$ and if $$xin[a,b]$$, then $$|f_n(x)-f(x)|leqslantfracvarepsilon2.

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\$

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, 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 $$sup_{[a,b]}|f(x)-f_n(x)| for all $$n>N$$. Thus for any $$xin [a,b]$$ and $$n>N$$ we have $$|f(x)-f_n(x)|^p. 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\$

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, 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 $$sup_{[a,b]}|f(x)-f_n(x)| for all $$n>N$$. Thus for any $$xin [a,b]$$ and $$n>N$$ we have $$|f(x)-f_n(x)|^p. 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

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.

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

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 $$e^x(1-n/2(exp(1/n) – exp(-1/n))) for $$n geq N$$ very much depends on $$x$$.

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

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 $$e^x(1-n/2(exp(1/n) – exp(-1/n))) for $$n geq N$$ very much depends on $$x$$.

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

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 $$|x-a|. Take some $$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.

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

$$x_0

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

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

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 $$|x-a|. Take some $$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.

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

$$x_0

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