Is Spivak Cheating?

Question

I'm trying to understand this proof in the Spivak's book Calculus on manifolds page 16:

When I read this I had a feeling that in this part Spivak was informal and intuitive, almost like "cheating" to have a more clean proof.

Of course, it has to have some formal argument behind, but I couldn't identify it. I think it has to do with limits of composition of functions, where one of the functions is $tx$.

Could anyone clarify this part?

Thanks

$tx$ is a vector. $\lambda$ and $\mu$ are linear, so $\lambda(tx) - \mu(tx) = t(\lambda(x)-\mu(x))$. — Daniel Fischer, Dec 06 '13 at 15:21
If I understand correctly, your question amounts to asking whether it is true that: If $\lim_{h \to 0} F(h) = 0$, then $\lim_{t \to 0} F(tx) = 0$. Yes? — Jesse Madnick, Dec 06 '13 at 15:21

Jesse Madnick · Accepted Answer · 2017-09-29T22:51:31.093

11

Suppose we know that $\displaystyle \lim_{h \to 0} F(h) = 0$. This means that for every $\epsilon > 0$, there exists $\delta > 0$ such that $$|h| < \delta \implies |F(h)| < \epsilon.$$

Fix $x \in \mathbb{R}^n$. If $|t| < \delta/|x|$, then $|tx| = |t||x| < \delta$, so $|F(tx)| < \epsilon$. This says exactly that $\displaystyle \lim_{t \to 0} F(tx) = 0$.

In your situation, take $$F(h) := \frac{|\lambda(h) - \mu(h)|}{|h|}.$$

Edit: If $\displaystyle \lim_{h \to h_0} F(h) = L$ and $\displaystyle \lim_{t \to t_0} g(t) = h_0$, then $\displaystyle \lim_{t \to t_0} F(g(t)) = L$.

Proof: Fix $\epsilon > 0$. By the first limit hypothesis, there exists $\delta_1 > 0$ such that $$|h - h_0| < \delta_1 \implies |F(h) - L| < \epsilon$$ By the second limit hypothesis, there exists $\delta_2 > 0$ such that $$|t - t_0| < \delta_2 \implies |g(t) - h_0| < \delta_1.$$ Therefore, if $|t - t_0| < \delta_2$, then $|F(g(t)) - L| < \epsilon$. This says exactly that $\lim_{t \to t_0} F(g(t)) = L$.

Remark: Note that no continuity hypotheses were necessary. A continuity hypothesis on $F$ would be needed to show instead that $\displaystyle \lim_{t \to t_0}F(g(t)) = F(h_0)$.

edited Sep 29 '17 at 22:51

answered Dec 06 '13 at 15:35

Jesse Madnick

31,524

thank you for your answer I would like a kind of "composition rule" for the limit to prove this, but your proof is really nice, thank you very much. – user108205 Dec 06 '13 at 15:44
In the final remark you said "A continuity hypothesis on $F$ would be needed to show instead that $\lim_{t \to t_0}F(g(t)) = F(h_0)$." to have the last equality true, we have to have both functions continuous or just $F$? – user108205 Dec 06 '13 at 16:42
Just $F$. ${}{}$ – Jesse Madnick Dec 06 '13 at 17:49
Very complete answer, great! Thanks a lot! – user108205 Dec 07 '13 at 10:10

score 7 · Answer 2 · answered Dec 06 '13 at 15:22

7

You fix $x \neq 0$ and $h=tx$ for $t \to 0$. You have proved that $$ \lim_{h \to 0} \frac{|\lambda (h)-\mu (h)|}{|h|}=0, $$ so $$ \lim_{t \to 0} \frac{|\lambda (tx)-\mu (tx)|}{|tx|}=0 $$ as well. But $\lambda (tx)=t \lambda (x)$ and $\mu (tx)=t \mu (x)$ by linearity, and you conclude.

answered Dec 06 '13 at 15:22

Siminore

35,136

1

I don't think this is what's confusing the OP. See my comment above. – Jesse Madnick Dec 06 '13 at 15:23
@JesseMadnick you're right. – user108205 Dec 06 '13 at 15:24
What I didn't understand is why the first limit you wrote is 0 – user108205 Dec 06 '13 at 15:25
1

Spivak proves it! The fraction is gretater than zero and smaller that a quantity that converges to zero. By comparison, the fration must tend to zero. By this should be trivial for you, if you want to read Spivak's book. – Siminore Dec 06 '13 at 15:26
One probably should distinguish between limit from the right and from the left, but basically you have the correct insight. – egreg Dec 06 '13 at 15:27
@user108205: Now I'm confused. Isn't your confusion regarding going from the first limit to the second? That is, in how the first limit being zero implies that the second limit is zero? Or am I misunderstanding? – Jesse Madnick Dec 06 '13 at 15:27
@Siminore yes I'm sorry I was confused, what I didn't understand was why the second limit you wrote is 0. – user108205 Dec 06 '13 at 15:28
@JesseMadnick yes sorry, see my last comment – user108205 Dec 06 '13 at 15:29
If $\lim_{x \to x_0} f(x)=\ell$ and $\lim_{t \to t_0} \phi(t)=x_0$ with $\phi(t) \neq x_0$ for $t$ close but different than $t_0$, then $\lim_{t \to t_0} f(\phi(t))=\ell$. – Siminore Dec 06 '13 at 15:30
Do you have any source where I can see this rule? – user108205 Dec 06 '13 at 15:35
http://math.stackexchange.com/questions/285667/how-to-prove-this-limit-composition-theorem – Siminore Dec 06 '13 at 15:44
Are there some requirements over $f$ and $\phi$? some of them has to be continuous to theorem holds? – user108205 Dec 06 '13 at 15:51
1

If they were continuous, why should we write limits in the assumptions? – Siminore Dec 06 '13 at 15:53

pppqqq · Answer 3 · 2013-12-06T17:03:04.807

4

The second chain of inequality shows that $$\lim _{h\to 0} \frac{|\lambda (h) - \mu (h)|}{|h|}=0.$$ A necessary condition for the convergence is that $\phi (t)\to 0$ if $t\to 0$ , where $\phi$ is the quotient calculated in $tx$. It's something like “$a_n \to a$ implies $a_{n_k}\to a$ for each subsequence $a_{n_k}$”.

edited Dec 06 '13 at 17:03

answered Dec 06 '13 at 15:23

pppqqq

2,150

William Vickery · Answer 4 · 2016-04-27T06:08:48.670

$\mu$ and $\lambda$ are linear transformations and $t$ is a scalar so $$ \mu(tx) - \lambda(tx) =t\mu(x) - t\lambda(x) = |t|(\mu(x) - \lambda(x)). $$ So $$ \left|\mu(tx) - \lambda(tx)\right| = |t|\left|\mu(x) - \lambda(x)\right|. $$ Also |tx| = |t||x| is a basic property of the euclidean norm, so $$ \left|\frac{\mu(tx) - \lambda(tx)}{tx}\right|=\frac{|t||\mu(x) - \lambda(x)|}{|t||x|} = \left|\frac{\mu(x) - \lambda(x)}{x}\right| $$ for $ t \neq 0$.

Spivak asserts $0 = \lim_{t \to 0} \left|\frac{\mu(tx) - \lambda(tx)}{tx}\right|$. So for all $\epsilon >0$ there exists a $\delta >0$ so that when $0<t<\delta$ $$ \epsilon > \left|\left| \frac{\mu(tx) - \lambda(tx)}{tx}\right| - 0\right| = \left| \frac{\mu(tx) - \lambda(tx)}{tx}\right| = \left|\frac{\mu(x) - \lambda(x)}{x}\right|. $$

Since $$ \left|\frac{\mu(x) - \lambda(x)}{x}\right| < \epsilon$$ for all $\epsilon > 0$ $$ \left|\frac{\mu(x) - \lambda(x)}{x}\right| = 0.$$

Since $|x| > 0$ then $\frac{1}{|x|} > 0$ so we must have $$ |\mu(x) - \lambda(x)| = 0$$ since there are no nilpotent elements in $\mathbb{R}$

What do the expressions in your first chain of equalities mean? You appear to be dividing a vector by a vector? — , Apr 26 '16 at 22:31

Is Spivak Cheating?

4 Answers4