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Let $\mu: [x,y]\to \mathbb{R}^n$ be a parametrized curve in $\mathbb{R}^n$ such that $\mu(x)=x_1$ and $\mu(y) = y_1$. Show that for any constant vector $v$ where $\lVert v \rVert=1$ then: $$(y_1-x_1) \dot\ v = \int^y_x \mu'(t) \dot\ v \ dt \le \int^y_x \lVert\mu'(t)\rVert dt.$$ Using this show that $$\lVert\mu(y)-\mu(x)\rVert\le \int^y_x \lVert\mu'(t)\rVert dt. $$

How can I prove this?

For the first part I know that I must use Cauchy-Schwarz inequality but I don't know how to apply it here. The second part is telling me the distance (or length) between $x_1$ and $y_1$ is less than the length (arc length) of $\mu$?

Pedro
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Lays
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2 Answers2

1

You're right. The first is Cauchy-Schwarz: Just integrate. For the second, pick the right unit vector $v$.

Pedro
  • 122,002
Ted Shifrin
  • 115,160
1

For the first part, Cauchy-Schwarz tells us that

$$\mu'(t) \cdot v \leq |\mu'(t) \cdot v| \leq ||\mu'(t)|| \cdot||v|| = ||\mu'(t)||$$

Since $||v|| =1$. Applying integrals, we get

$$ (y_1 - x_1) \cdot v = \int_{x}^{y} \mu'(t) \cdot v \leq \int_{x}^{y} ||\mu'(t)||$$

For the second part, the hint above is good. If you cannot figure it out, the solutions is below (highlight to see the math):

Note that you can take the absolute value of both sides and preserve the inequality, since the right-hand side is already positive. But

$$\color{white}{ ||(y_1 - x_1) \cdot v|| = ||(y_1 - x_1)|| \cdot ||v|| \cdot \cos \theta }$$

So choose a unit length $v$ for which $\cos \theta = 1$

Elchanan Solomon
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  • Why spoil another user's answer, which guides the OP rather than spoon-feeding with a full solution? It is true one is free to answer what one seems fit, but for homework problems like this, and with an already hinting answer, why give it away? – Pedro Sep 03 '13 at 02:21
  • The OP already said that he needed to use Cauchy-Schwarz, so I did not see the benefit in any hint repeating that fact. You are correct, though, that I should not have expounded upon the second hint. Let me hide that now. – Elchanan Solomon Sep 03 '13 at 02:25
  • Use \lVert and \rVert for a nicer looking $\lVert x\lVert$. – Pedro Sep 03 '13 at 02:41
  • @IsaacSolomon Thank you! I understand clearly now how to do part 1. But for part 2 I am not so sure how you got that? – Lays Sep 03 '13 at 02:47
  • @Lays what exactly are you unsure about? – Elchanan Solomon Sep 03 '13 at 11:26
  • I understand if I take absolute values it will still preserve angles but why introduce $v$? – Lays Sep 03 '13 at 18:42
  • The proof above takes advantage of the Cauchy-Schwarz inequality, so you introduce $v$ because Cauchy-Schwarz involves the dot product of two different vectors. Once you get the first inequality, you want to replace the left-hand side with $||\mu(y) - \mu(x)||$, which requires you to take the absolute value of $(y_1 - x_1) \cdot v$, and then make a clever choice of $v$ that gives equality in the Cauchy-Schwarz inequality. – Elchanan Solomon Sep 04 '13 at 07:34
  • So if I take the absolute value for $(y_1 - x_1) \cdot v$ then I will get $||\mu(y) - \mu(x)|| = ||\mu (t)||\cdot v$ then if I choose $v = \frac{\mu(a) - \mu(b)}{||\mu(a) - \mu(b)||}$ is that correct? – Lays Sep 04 '13 at 18:06
  • What you mean to say is: the absolute value of $(y_1 - x_1) \cdot v = \lvert y_1 - x_1 \rvert \lvert v \rvert \cos \theta$. Then if we choose the $v$ you have defined (assuming $x_1 \neq y_1$), then the norm of $v$ is $1$, and $\cos \theta$ will also be $1$, and the proof is complete. – Elchanan Solomon Sep 04 '13 at 21:18
  • Thanks, sorry if I am troubling, but just to make sure I follow you.

    We start with $(y_1 - x_1) \cdot v$ which by definition equals $ \lvert y_1 - x_1 \rvert \lvert v \rvert \cos \theta$. Now if we take $v = \frac{\mu(a) - \mu(b)}{||\mu(a) - \mu(b)||}$ where $a \ne b$ how can I construct it to complete the proof that $\lVert\mu(y)-\mu(x)\rVert\le \int^y_x \lVert\mu'(t)\rVert dt.$?

     – Lays Sep 05 '13 at 00:06
  • You mean the absolute value of $(y_1 - x_1) \cdot v$ equals that. If you take $v$ as above, then $|y_1 - x_1| |v| \cos \theta = |y_1 - x_1|$. Then simply observe that $|(y_1 - x_1) \cdot v| \leq \int_{x}^{y} |\mu'(t)|$ from part (1). – Elchanan Solomon Sep 05 '13 at 02:14