An additive map that is not a linear transformation over $mathbb{R}$, when $mathbb{R}$ is considered as a $mathbb{Q}$-vector space

I am looking for an example of an additive map that is not a linear transformation over $\mathbb{R}$, when $\mathbb{R}$ is considered as a $\mathbb{Q}$-vector space. I mean, I want to find an example of a map $T:\mathbb{R}\rightarrow\mathbb{R}$ such that $T(u+v)=T(u)+T(v)$ for all $u,v\in \mathbb{R}$, but $T(\alpha v)=\alpha T(u)$ is not true for all $\alpha \in\mathbb{R}$.

Thanks for your kindly help.

Answer

Let $\{r_\alpha\}$ be a Hamel basis of $\Bbb R$ over $\Bbb Q$. Let $\phi$ map $x$ to $c_{\alpha_1}+\cdots+c_{\alpha_k}$, where the (unique) basis representation of $x$ is $c_{\alpha_1}r_{\alpha_1}+\cdots+c_{\alpha_k}r_{\alpha_k}$. Then $\phi(x+y)=\phi(x)+\phi(y)$, but takes on only rational values.

If $\phi(\alpha v)=\alpha\phi(v)$ for all $\alpha$, $v$ in $\Bbb R$, then $\phi$ would be onto.
As this isn't the case, $\phi$ is not $\Bbb R$-linear.

It is $\Bbb Q$-linear, though. In fact, any additive map would automatically be $\Bbb Q$-linear.

As far as I know, you need the axiom of choice to construct a function of this type (?).

calculus - Find $lim _{ nrightarrow infty }{ sum _{ k=1 }^{ n }{ frac { sqrt { k } }{ { n }^{ frac { 3 }{ 2 } } } } } $

Need help find the limit of $\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ \frac { \sqrt { k } }{ { n }^{ \frac { 3 }{ 2 } } } } } $

Now my intuition is that using Stolz-Cesaro

$\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ \frac { \sqrt { k } }{ { n }^{ \frac { 3 }{ 2 } } } } }=\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } \sum _{ k=1 }^{ n }{ \sqrt { \frac { k }{ n } } } =1 } $

Is it correct or not?

Answer

how about using Riemann sums??
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{\sqrt k}{n^{3/2}} = \lim_{n\to\infty}\frac 1 n \sum_{k=1}^n \sqrt{\frac k n} = \int_0^1 \sqrt x dx $$

abstract algebra - Linearly disjoint fields

We say that two field $E,F$, extending the same base field $K$, are linearly disjoint if every finite subset of $E$ that is $K$-linearly independent is also $F$-linearly independent.

Suppose $K = \mathbb{Q}$. Is this definition equivalent to say that $E \cap F = \mathbb{Q}$? And if so, why?

My attempt: Assuming that the extensions $E/\mathbb{Q}$, $F/\mathbb{Q}$ are finite, I tried using the primitive element theorem, so that $E=\mathbb{Q}(\alpha)$ and $F=\mathbb{Q}(\beta)$, for some $\alpha,\beta$ algebraic. Then the elements of these fields are just polynomials in these numbers, but from here i was not able to conclude.

Is is even true if the extensions are not finite?

Thanks in advance!

Answer

$\newcommand{\Q}{\mathbb{Q}}$No, it is not equivalent.

As a possibly typical example, take $K = \Q$, $E = \Q(\omega \alpha)$, $F = \Q(\alpha)$, where $\alpha = \sqrt[3]{2}$ and $\omega$ is a primitive third root of unity.

We have $E \cap F = K$, but while $1, \omega \alpha, \omega^{2} \alpha^{2} \in E$ are independent over $K$, you have
$$
1 + \frac{\alpha^{2}}{2}( \omega \alpha) + \frac{\alpha}{2} (\omega^{2} \alpha^{2}) = 1 + \omega + \omega^{2} = 0,
$$

so they are not independent over $F$.

field theory - Doubt in Lang's proof that normal extensions remain normal under lifting (Theorem V.3.4, *Algebra*)

Theorem 3.4. Normal extensions remain normal under lifting. If $K \supset E \supset k$ and $K$ is normal over $k$, then $K$ is normal over $E$. If $K_1$, $K_2$ are normal over $k$ and are contained in some field $L$, then $K_1 K_2$ is normal over $k$, and so is $K_1 \cap K_2$.

Proof. For our first assertion, let $K$ be normal over $k$, let $F$ be any extension of $k$, and assume $K$, $F$ are contained in some bigger field. Let $\sigma$ be an embedding of $KF$ over $F$ (in $F^*$). Then $\sigma$ induces the identity on $F$, hence on $k$, and by hypothesis its restriction to $K$ maps $K$ into itself. We get $(KF)^\sigma = K^\sigma F^\sigma = KF$ whence $KF$ is normal over $F$.

Lang proves the above theorem in his book Algebra. The only first part of the proof seems to be weird, namely that "lifting respects normal extensions".

Indeed, he has shown that any embedding $\sigma$ of $KF$ in $F^a$ over $F$ gives an automorphism. And this is one of the equivalent definitions of normal extensions.

But we also have to show that $KF$ is algebraic over $F,$ which I do not see in his proof (because the definition of normal extension requires an algebraic extension with certain properties).

How to show that $KF$ is algebraic over $F$?

Would be very grateful for any help!

calculus - A sine integral $int_0^{infty} left(frac{sin x }{x }right)^n,mathrm{d}x$

The following question comes from Some integral with sine post
$$\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$$
but now I'd be curious to know how to deal with it by methods of complex analysis.
Some suggestions, hints? Thanks!!!

Sis.

Answer

Here's another approach.

We have

$$\begin{eqnarray*}
\int_0^\infty dx\, \left(\frac{\sin x}{x}\right)^n
&=& \lim_{\epsilon\to 0^+}
\frac{1}{2} \int_{-\infty}^\infty dx\,
\left(\frac{\sin x}{x-i\epsilon}\right)^n \\
&=& \lim_{\epsilon\to 0^+}
\frac{1}{2} \int_{-\infty}^\infty dx\,
\frac{1}{(x-i\epsilon)^n}
\left(\frac{e^{i x}-e^{-i x}}{2i}\right)^n \\
&=& \lim_{\epsilon\to 0^+}

\frac{1}{2} \frac{1}{(2i)^n} \int_{-\infty}^\infty dx\,
\frac{1}{(x-i\epsilon)^n}
\sum_{k=0}^n (-1)^k {n \choose k} e^{i x(n-2k)} \\
&=& \lim_{\epsilon\to 0^+}
\frac{1}{2} \frac{1}{(2i)^n}
\sum_{k=0}^n (-1)^k {n \choose k}
\int_{-\infty}^\infty dx\, \frac{e^{i x(n-2k)}}{(x-i\epsilon)^n}.
\end{eqnarray*}$$
If $n-2k \ge 0$ we close the contour in the upper half-plane and pick up the residue at $x=i\epsilon$.
Otherwise we close the contour in the lower half-plane and pick up no residues.

The upper limit of the sum is thus $\lfloor n/2\rfloor$.
Therefore, using the Cauchy differentiation formula, we find
$$\begin{eqnarray*}
\int_0^\infty dx\, \left(\frac{\sin x}{x}\right)^n
&=& \frac{1}{2} \frac{1}{(2i)^n}
\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k}
\frac{2\pi i}{(n-1)!}
\left.\frac{d^{n-1}}{d x^{n-1}} e^{i x(n-2k)}\right|_{x=0} \\
&=& \frac{1}{2} \frac{1}{(2i)^n}
\sum_{k=0}^{\lfloor n/2\rfloor}

(-1)^k {n \choose k}
\frac{2\pi i}{(n-1)!} (i(n-2k))^{n-1} \\
&=& \frac{\pi}{2^n (n-1)!}
\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}.
\end{eqnarray*}$$
The sum can be written in terms of the hypergeometric function but the result is not particularly enlightening.

Is this intuition behind product manifolds correct?

I've been studying differential geometry on Spivak's books and recently I proved that the cartesian product of manifolds is another manifold. Right, however, what's the intuition behind this? I've tried to develop some intuition, but I'm not sure it's correct.

For instance, we can say that the cylinder $C$ is the product of the circle and the real line, in other words $C = S^1 \times \mathbb{R}$. My intuition says that we do define this way because identifying a point on a cylinder is equivalent to identifying a point on a circle and then the position of this point on a line. In other words, describing one point on a cylinder should be equivalent to describing one point on a circle plus one point on a line.

Another example, the thorus given by $T = S^1 \times S^1$, again what's the intuition behind it? My idea is: given that we can describe the torus as the result of rotating a circle around some axes we would describe a point on the torus to be one point on the circle being rotated plus one point on the circle that represents the trajectory of that particular point during the rotation.

So in general, my intuition is that given $r$ manifolds $M_1, \dots, M_r$ their product is a manifold that each point can be described by one point on each manifold in a particular way depending on the case we are considering.

Is this intuition correct? Is this the way we should think about product manifolds?

Thanks in advance for your help.

Answer

For the product manifold there is a different explanation.

Take the example $S^1\times\mathbb R$. This means that at each point of the circle $S^1$ you attach the whole real line!! So in this way, going round the circle and attaching everywhere a real line you get a cylinder.

For the other example : $\mathbb T^2=S^1\times S^1$. At each point of a circle attach another circle (orthogonal to the 1st circle). Then, if for all points of the circle you attach an orthogonal circle, you are going to get the torus!

It's pretty fascinating how intuition works here!

integration - If ${f_n}$ uniformly integral, then ${ f_n - f }$ uniformly integrable?

Let $\{f_n\}$ be uniformly integrable, on space $(\Omega, F, P)$. $f_n \rightarrow f$ in measure.

The notes I was reading then stated two facts i do not follow:

• By Fatou's $\int_\Omega |f| dP \le \sup_n \int_\Omega |f_n | dP.$

we should have "$\liminf |f_n|$" on the LHS. How is this implied?

• $\{f_n -f\}$ is uniformly integrable.

I tried to show $\sup_n \int_{|f_n-f|>N} |f_n-f| dP$ can be bounded but could not split the integral into easier pieces. How does this follow?

Answer

If $f_n \to f$ in measure, then there is a subsequence $\{f_{n_k}\}$ with $f_{n_k}(x) \to f(x)$ almost everywhere. Fatou's lemma gives you

$$\int_\Omega |f| \, dP = \int_\Omega \lim_{k \to \infty} |f_{n_k}| \, dP \le \lim_{k \to \infty} \int_\Omega |f_{n_k}| \, dP \le \sup_n \int_\Omega |f_n| \, dP.$$

Uniform integrability means two things:

• $\displaystyle \sup_n \int_\Omega |f_n| \, dP < \infty$, and




• $\forall \epsilon > 0 \ \exists \delta > 0$ $P(A) < \delta \implies \displaystyle \sup_n \int_A |f_n| \, dP < \epsilon$.

The remark about Fatou tells you that $\displaystyle \int_\Omega |f| \, dP < \infty$. In particular, for any $\epsilon > 0$ there exists $\delta > 0$ with the property that $P(A) < \delta$ implies $\displaystyle \int_A |f| \, dP < \epsilon.$

Now work with $\epsilon$ and $\delta$ and the fact that $$\int_A |f_n - f| \, dP \le \int_A |f_n| \, dP + \int_A |f| \, dP$$ to show $\{|f_n - f|\}$ is uniformly integrable.