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

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’}|.

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)\$?

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)|

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|

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.

## 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)\$?

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)|

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|

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.

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

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!

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

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|

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|

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|

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

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

$$|x-z|

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## 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: 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)\$?

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)|

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|

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.

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

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 $$d(x,y) leq d(x,x_n) + d(x_n,y) = 2d(x,x_n) which is impossible. Thus $$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)?

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

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 $$left = left$$.

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.

$$begin{cases}x=t^3\y=t^2end{cases}$$

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 $$y = (x^{1/3})^2$$.

## Math Genius: Continuity of a topological conjugacy \$h\$

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.