Math Genius: $f(x,y)=arcsin frac{x}{y}$ is continuous but not uniformly continuous in its domain

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I need to prove that $f(x,y)=arcsin frac{x}{y}$ is continuous, but not uniformly continuous on its domain.
I noticed that the domain of the function is $D_f={(x,y)|-yleq x leq y$ if $y>0$, and $yleq x leq -y$ if $y<0}.$
I started to prove the continuity by the epsilon-delta deffinition, but I’m stuck at proving that $|arcsin frac{x}{y} – arcsin frac{x’}{y’}|<epsilon$.

To show that the function $f$ is not unifromly continous first recall that $operatorname{arcsin} 1 =tfrac{pi}2$ and $operatorname{arcsin} tfrac 12 =tfrac{pi}6$. Thus for each natural $n$ we have $fleft(tfrac 1n, tfrac 1nright)= tfrac{pi}2$ and $fleft(tfrac 1n, tfrac 2nright)= tfrac{pi}6$, whereas $left|left(tfrac 1n, tfrac 1nright)- left(tfrac 1n, tfrac 2nright)right|=tfrac 1n$.

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Math Genius: Why does $f(x,y)= frac{xy^2}{x^2+y^4}$ have different limits when approaching $(0,0)$ along straight lines vs. along the curve $(1/t^2,1/t)$?

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I have a stupid question about continuity in higher dimensions.

There are maps, for example, $f(x,y)=frac{xy^2}{x^2+y^4}$, when $(x,y)neq (0,0)$ and $f(x,y)=(0,0)$ when $(x,y)=(0,0)$, when we approach $(0,0)$ along every straight line, the limit of the function is $0$, but when along a curve, for example $(frac{1}{t^2},frac{1}{t})$, the limit of $f$ is not $0$.

But, it feels like all the straight lines can cover a neighbourhood of $(0,0)$, so every point on a curve is also on a different straight line. Why is it that when the same points are arranged in a different way the limit changes?

The reason is that the function is approaching zero along each straight line at a different speed, depending on the line. So, for $f$ to be, say, less than $1/10$ along one of the lines, you need to be within a distance $1$ from the origin, whereas for a different line you need to be within a distance of $1/2$ units, etc. It is then perfectly possible that if you approach $(0,0)$ along a curve that is transversal to all those lines, at the points of intersections with the lines your function is all the time equal to $1/10$, which makes the limit along the curve equal $1/10$. This does not contradict the fact that along all the lines the limit is zero.

This has something to do with the uniform continuity around 0. Given a fixed $epsilon > 0$, if there exist an universal constant $delta_u$ such that

$$|(x,y)|<delta_u implies |f(x,y)|<epsilon$$

then your intuition is right: along any curve go to 0, $f$ will have limit 0.

However, for this example the origin is not uniformly continuous. Along lines $y=ax$, $a inmathbb{R}$ we can define a class of fucntions $f_a(r) = frac{a^2 x}{1+a^4 x^2}$. Suppose the universal constant $delta_u$ exist, then for every $a$ we must have

$$ |x|<delta_u implies |f_a(x)|<epsilon $$

However, let $a=frac{1}{sqrt{delta_u}}$ we have $f_a(delta_u) = 1/2$. Therefore, above argument cannot be correct, and the origin is not uniform continuous with respect to ${f_a(x)}$.

For the same reason that traveling from New York to Florida takes a few hours if you fly straight south, but takes a lot more if you choose to go first to Los Angeles.

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Math Genius: Why does $f(x,y)= frac{xy^2}{x^2+y^4}$ have different limits when approaching $(0,0)$ along straight lines vs. along the curve $(1/t^2,1/t)$?

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I have a stupid question about continuity in higher dimensions.

There are maps, for example, $f(x,y)=frac{xy^2}{x^2+y^4}$, when $(x,y)neq (0,0)$ and $f(x,y)=(0,0)$ when $(x,y)=(0,0)$, when we approach $(0,0)$ along every straight line, the limit of the function is $0$, but when along a curve, for example $(frac{1}{t^2},frac{1}{t})$, the limit of $f$ is not $0$.

But, it feels like all the straight lines can cover a neighbourhood of $(0,0)$, so every point on a curve is also on a different straight line. Why is it that when the same points are arranged in a different way the limit changes?

The reason is that the function is approaching zero along each straight line at a different speed, depending on the line. So, for $f$ to be, say, less than $1/10$ along one of the lines, you need to be within a distance $1$ from the origin, whereas for a different line you need to be within a distance of $1/2$ units, etc. It is then perfectly possible that if you approach $(0,0)$ along a curve that is transversal to all those lines, at the points of intersections with the lines your function is all the time equal to $1/10$, which makes the limit along the curve equal $1/10$. This does not contradict the fact that along all the lines the limit is zero.

This has something to do with the uniform continuity around 0. Given a fixed $epsilon > 0$, if there exist an universal constant $delta_u$ such that

$$|(x,y)|<delta_u implies |f(x,y)|<epsilon$$

then your intuition is right: along any curve go to 0, $f$ will have limit 0.

However, for this example the origin is not uniformly continuous. Along lines $y=ax$, $a inmathbb{R}$ we can define a class of fucntions $f_a(r) = frac{a^2 x}{1+a^4 x^2}$. Suppose the universal constant $delta_u$ exist, then for every $a$ we must have

$$ |x|<delta_u implies |f_a(x)|<epsilon $$

However, let $a=frac{1}{sqrt{delta_u}}$ we have $f_a(delta_u) = 1/2$. Therefore, above argument cannot be correct, and the origin is not uniform continuous with respect to ${f_a(x)}$.

For the same reason that traveling from New York to Florida takes a few hours if you fly straight south, but takes a lot more if you choose to go first to Los Angeles.

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Math Genius: Continuity of a recursive piece-wise function.

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I’m a student studying math, and I’m going through some old exam problems and I have come across a set of questions that ask me to decide where a given piece-wise function is continuous. The function in question is.

$$f(x)= begin{cases}
|x-1| & text{if }x geq 0 \
f(-x) & text{if }x <0 \
end{cases}
$$

This function naturally is causing me much confusion. The function seems it will trail of into a infinite string of functions in functions. However If I had to guess its continuity I’d hazard intuitively there appears to be no reason $f(x)$ is not defined across $mathbb{R}$. Am I right to think this? Is there a rigorous theorem I should be using? And how do I manage this recursive second condition?

Any help would be much appreciated!

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Math Genius: Continuity of a function at a point in $mathbb{R}_+$

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I am given a function where
$$f:Ssubset mathbb{R}_+rightarrow mathbb{R} $$

$$f(x)=frac{sqrt{x}-1}{x-1}$$

I need to use the following definition to prove that $f$ is continuous at $zin S$;

$f(x)$ converges to the limit, L, as $x$ tends to $z$, for every $epsilon>0$ , there exists $delta>0$ such that:

$|x-z|<delta Rightarrow |f(x)-L|<epsilon $

I also have the following information that I need to use to prove continuity at $zin S$:
$$|frac{1}{sqrt{x}+1}-frac{1}{sqrt{z}+1}|<|sqrt{z}-sqrt{x}|$$

I am also given a hint that $sqrt{z}ge0$ which implies $sqrt{x}+sqrt{z}ge sqrt{x}-sqrt{z}$

What I have done so far is:

$lim_{x to z} f(x) = frac{sqrt{z}-1}{z-1}$

$|x-z|<delta Rightarrow |frac{sqrt{x}-1}{x-1}-frac{sqrt{z}-1}{z-1}|<epsilon $

At this point, I’m not sure how to incorporate the information into proving that $f$ is continuous at $zin S$

Forgive me, I am not the brightest mathematician, any help or hints to point me in the right direction would be appreciated it

Using the fact that if $x>0$, then $x-1=(sqrt{x}-1)(sqrt{x}+1)$

I need to show to show that for a given $epsilon>0$, there exists a $delta$ such that:

$$|x-z|<delta Rightarrow|frac{1}{sqrt{x}+1}-frac{1}{sqrt{z}+1}|<epsilon$$

Since

$zge 0$, it follows that; $sqrt{x}-sqrt{z}lesqrt{x}-sqrt{z}$

[By multiplying the inequality on both sides by $sqrt{x}-sqrt{z}$]

Note: I have left out the steps of derivation to keep it short. But by multiplying on both sides we arrive at the following:

$$(sqrt{x}-sqrt{z})^2<x-z$$

We know that
$$|x-z|<delta$$
Then we use the information to show that
$$|frac{1}{sqrt{x}+1}-frac{1}{sqrt{z}+1}|<|sqrt{x}-sqrt{z}|<sqrt{delta}$$
Then by choosing a $delta=epsilon^2$ , we have proven that $f$ is continuous at $zin S$ such that:

$$|x-z|<delta Rightarrow|frac{1}{sqrt{x}+1}-frac{1}{sqrt{z}+1}|<sqrt{epsilon^2}=epsilon$$

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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: Why does $f(x,y)= frac{xy^2}{x^2+y^4}$ have different limits when approaching $(0,0)$ along straight lines vs. along the curve $(1/t^2,1/t)$?

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I have a stupid question about continuity in higher dimensions.

There are maps, for example, $f(x,y)=frac{xy^2}{x^2+y^4}$, when $(x,y)neq (0,0)$ and $f(x,y)=(0,0)$ when $(x,y)=(0,0)$, when we approach $(0,0)$ along every straight line, the limit of the function is $0$, but when along a curve, for example $(frac{1}{t^2},frac{1}{t})$, the limit of $f$ is not $0$.

But, it feels like all the straight lines can cover a neighbourhood of $(0,0)$, so every point on a curve is also on a different straight line. Why is it that when the same points are arranged in a different way the limit changes?

The reason is that the function is approaching zero along each straight line at a different speed, depending on the line. So, for $f$ to be, say, less than $1/10$ along one of the lines, you need to be within a distance $1$ from the origin, whereas for a different line you need to be within a distance of $1/2$ units, etc. It is then perfectly possible that if you approach $(0,0)$ along a curve that is transversal to all those lines, at the points of intersections with the lines your function is all the time equal to $1/10$, which makes the limit along the curve equal $1/10$. This does not contradict the fact that along all the lines the limit is zero.

This has something to do with the uniform continuity around 0. Given a fixed $epsilon > 0$, if there exist an universal constant $delta_u$ such that

$$|(x,y)|<delta_u implies |f(x,y)|<epsilon$$

then your intuition is right: along any curve go to 0, $f$ will have limit 0.

However, for this example the origin is not uniformly continuous. Along lines $y=ax$, $a inmathbb{R}$ we can define a class of fucntions $f_a(r) = frac{a^2 x}{1+a^4 x^2}$. Suppose the universal constant $delta_u$ exist, then for every $a$ we must have

$$ |x|<delta_u implies |f_a(x)|<epsilon $$

However, let $a=frac{1}{sqrt{delta_u}}$ we have $f_a(delta_u) = 1/2$. Therefore, above argument cannot be correct, and the origin is not uniform continuous with respect to ${f_a(x)}$.

For the same reason that traveling from New York to Florida takes a few hours if you fly straight south, but takes a lot more if you choose to go first to Los Angeles.

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Math Genius: Show that if $X$ is compact metrizable then $C(X)$ is separable.

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Let $X$ be a metrizable compact space. I want to show that $C(X, mathbb{C})$ is separable in the uniform topology.

Attempt: By our assumption, $X$ is separable, so we can pick a countable dense subset ${x_n:n geq 1}$. Let $d$ be a metric inducing the topology on $X$. Define for $n geq 1$

$$d_n: X to mathbb{R}: x mapsto d(x,x_n)$$

Let $$mathcal{A}= bigcup_{n=1}^infty mathbb{C}[d_1, dots, d_n]$$
where $mathbb{C}[d_1, dots, d_n]$ is the set of complex polynomials in the functions $d_1, dots, d_n$.

It is easily checked that $mathcal{A}$ is a subalgebra of $C(X, mathbb{C})$, since $mathcal{A}$ is closed under addition, multiplication and scalar multiplication. Moreover, $1 in mathbb{C}[d_1]subseteq mathcal{A}$ so the algebra is unital. Clearly $mathcal{A}$ is closed under complex conjugation, since the functions $d_1, d_2, dots$ are all real-valued. We know that $mathcal{A}$ separates the functions of $C(X, mathbb{C})$:

If $x neq y$ with $x, y in X$. Use density to choose $n geq 1$ with $d(x_n,x) < d(x,y)/2$. Then we have $d(x_n,x) neq d(x_n,y)$. Otherwise $d(x_n,x) = d(x_n,y)$ and we get $d(x,y) leq d(x,x_n) + d(x_n,y) = 2d(x,x_n)<d(x,y)$ which is impossible. Thus $d_n(x) neq d_n(y)$ so our algebra separates the points.

By Stone-Weierstrass, $mathcal{A}$ is dense in $C(X, mathbb{C})$. Let $D$ be a countable dense subset of $mathbb{C}$, for example $D= mathbb{Q}+ i mathbb{Q}$. Any element of $mathcal{A}$ can be approximated by an element in $$mathcal{B}:= bigcup_{n=1}^infty D[d_1, dots, d_n]$$ and we conclude that $mathcal{B}$ is dense in $C(X, mathbb{C})$. Clearly $mathcal{B}$ is countable, and thus we conclude the proof. $quad square$

Is this proof correct?

The idea of the proof is fine, and standard, I think. Maybe you might want to add details about why your $mathcal{B}$ is dense, so why replacing the coefficients of the members of the algebra by a dense subset is OK. Maybe you’re relying on an earlier lemma? The countability might warrant a few words too, depending on the target audience (how much set theory do they know)?

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Math Genius: How does $(t^3, t^2)$ represent $x^{2/3}$?

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I just learnt about this parameterization and somehow I am not being able to wrap my head around it. How on earth is the curve $left<x, x^{2/3}right> = left<t^3, t^2right>$.

I got this from Claudio Arezzo’s class on differential geometry and I have verified it manually but I still can’t believe it / comprehend it.

If you make the variable substitution $x=t^3$ in the formula $langle x, x^{2/3} rangle$ you get
$$langle t^3, (t^3)^{2/3} rangle = langle t^3,t^2 rangle
$$

You can also do this in the reverse direction, substitute $x=t^{1/3}$ into the formula $langle t^3,t^2rangle$ and you get
$$langle (x^{1/3})^3, (x^{1/3})^2 rangle = langle x,x^{2/3} rangle
$$

This tells you that, as subsets of the coordinate plane, we have an equation
$${(x,x^{2/3}) mid x in mathbb R } = {(t^3,t^2) mid t in mathbb R}
$$

which is what is really meant by the not-really-very-rigorous equation $ langle x,x^{2/3}rangle = langle t^3,t^2 rangle$.

And by the way, it should not really be a big surprise that one can parameterize geometric shapes using different parameters. Just as an example, the line $x+y=1$ can be parameterized as $langle x,1-xrangle$ or as $langle 1-y,y rangle$ as well as in multiple other ways.

Start with the parametric graph
$$
begin{cases}x=t^3\y=t^2end{cases}
$$

We want to eliminate the parameter $t$ and just get $y=f(x)$. To do this:

Solve $x=t^3$ for $t$ to get $t = x^{1/3}$. Then substitute this
into $y=t^2$ to get $y = (x^{1/3})^2$.

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Math Genius: Continuity of a topological conjugacy $h$

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The system $x’=Ax$ is an attractor. Let $h$ be defined by

$$h(0)=0 qquad h(x)=e^{t_x}e^{t_xA}x$$
where $t_x$ is the real number such that $q(e^{t_x A}x)=1$ and $q(x)=int_0^{infty}langle e^{tA}x,e^{tA}xrangle,dt$.

The book says that $h(cdot)$ is continuous and is the topological conjugacy of $x’=Ax$ and $x’=-x$.

Knowing that $t_x$ is $C^{infty}$ in $mathbb{R}^nsetminus{0}$ I managed to show that $h|_{mathbb{R}^nsetminus{0}}$ is continuous. But I’m not able to show that this function is continuous at 0.

You first should show that there exist $a,b>0$ such that
$$
a|x|^2le q(x)le b|x|^2.
$$

This follows easily from having an attractor (use the eigenvalues of $A$).

Now observe that
$$
begin{split}
q(h(x))
&=int_0^{infty}langle e^{tA}e^{t_x}e^{t_xA}x,e^{tA}e^{t_x}e^{t_xA}xrangle,dt\ &=e^{2t_x}int_0^{infty}langle e^{tA}e^{t_xA}x,e^{tA}e^{t_xA}xrangle,dt\
&=e^{2t_x}q(e^{t_xA}x)=e^{2t_x}.
end{split}
$$

Therefore,
$$
a|h(x)|^2le e^{2t_x}le b|h(x)|^2.
$$

On the other hand, since
$$
1=q(e^{t_xA}x) =int_0^{infty}| e^{(t+t_x)A}x|^2,dt= int_{t_x}^{infty}| e^{tA}x|^2,dt,
$$

we find that if $xto0$, then $t_xto-infty$. It follows from $$a|h(x)|^2le e^{2t_x}$$ that $h(x)to0$ when $xto0$. The other inequality can be used for the inverse.

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