Let $\gamma$ be a curve such that $\gamma(a)=p$ and $\gamma(b)=q$ and let $u$ be a unit vector.
A) Prove $\gamma'(t) \cdot u$ $\leq$ $\|\gamma’(t)\|$
For this one I use the properties of dot product, noting that $u=\frac{\gamma’}{\|\gamma\|}$ but I always end up just showing that the quantities are equal.
B) Use path integrals to show ($\gamma(b)-\gamma(a)) \cdot u$ $\leq$ $\int_{a}^{b} \|\gamma'(t)\| dt$
For this one I observe that $u=\frac{(q-p)}{\|(q-p)\|}$ and sloppily reach a conclusion.
C) Prove the length of gamma from $p$ to $q$ is larger than or equal to the length of the straight line between $p$ and $q$.
This one is intuitively obvious, but for the proof I observe the straight line dist is just $\|p-q\|$ and the length of gamma is the arc length. I had myself convinced I had this part the other day, but now I’m confused.
Hopefully someone can make this all more clear for me. I apologize for the terrible (lack of) formatting
Thx
Nick
Best Answer
Here are three hints.
A) Use the Cauchy-Schwarz inequality $|\langle v, w \rangle|\leq \|v\| \|w\|$. What do you choose for $v$ and $w$?
B) Use the fact that for two real functions $f$ and $g$ with $f\leq g$, one has $$ \int_a^b f(t)\,dt \leq \int_a^b g(t)\,dt. $$ What $f$ and $g$ do you choose?
C) You have now proven the inequality $$(\gamma(b)-\gamma(a))\cdot u \leq \int_a^b \|\gamma'(t)\|\,dt$$ for every unit vector $u$. Note that the right hand side $\int_a^b \|\gamma'(t)\|\,dt$ is the length of $\gamma$. Choose an appropriate $u$ to show $\|p-q\|$ is smaller than the length of $\gamma$.