The technique for proving the triangle inequality Spivak used was interesting, basically, if x,y are linearly independent then, for all \lambda \in \mathbf{R}
0 \lt \|x - \lambda y\|^2 = \lambda^2 \|y\|^2 - 2\lambda \langle x,y\rangle + \|x\|^2
Since the quadratic has no real solution we must have b^2 - 4ac \lt 0 or
4\langle x,y\rangle^2 - 4\|y\|^2\|x\|^2 \lt 0
Which implies \langle x,y\rangle \lt \|x\|\|y\| as desired.
Intuitively taxicab distance must be longer, since it constrains you from going diagonally. For a proof, take x = x_ie^i then use the triangle inequality to get |x| \le \sum_{i=1}^n |x_i| (since |e_i|=1).
In Theorem 1-1 (3) the critical step is Cauchy Swartz, in which equality holds if and only if x = \lambda y for some \lambda.
This is just the triangle inequality (Theorem 1-1 (3)) on (-y), i.e. |x - y| = |x + (-y)| \le |x| + |-y| = |x| + |y|
WOLG assume |x| \ge |y|, then |x| - |y| \le |x-y| is a rearrangement of the triangle inequality, since |x| = |(x - y) + y| \le |x-y| + |y|
Obvious consequence of Theorem 1-1 (3).
|z-x| = |(z-y) + (y-x)| \le |z-y| + |y-x|
Exactly the same argument as before.
The assumption “\int_a^b f - \lambda g \ne 0 for all \lambda” and “f - \lambda g \ne 0 for all \lambda” are not the same in general (consider where f and g differ on a set of measure zero, e.g. a single point.) for continuous functions they are the same though, since any nonzero point contributes measure for continuous functions.
There’s a natural isomorphism between vectors in \mathbf{R}^n and functions on [0,1] that looks something like this (too lazy to write it down explicitly)
By “isomorphism” I mean that if \phi : \mathbf{R}^n \to C[0,1] is our isomorphism and x,y \in \mathbf{R}^n, then \langle x,y\rangle = \langle \phi(x), \phi(y)\rangle or more explicitly \sum_{i=1}^n x_iy_i = \int_0^1 \phi[x](t)\phi[y](t)dt
Trivial from the polarization identity (Theorem 1-2 (5)) but still profound. You can check angles are preserved by checking distances get preserved!
|Tx| = |x| = 0 if and only if x = 0 by the definiteness of inner products, likewise for y = Tx we have |T^{-1}y| = |x| = |Tx| = |y| meaning T^{-1} is also norm preserving.
\angle(Tx,Ty) = \arccos \frac{\langle Tx,Ty\rangle}{|Tx||Ty|} = \arccos \frac{\langle x,y\rangle}{|x||y|} = \angle(x,y)
\angle(Tx,Ty) = \arccos \frac{\langle Tx,Ty\rangle}{|Tx||Ty|} = \arccos \frac{\lambda^2\langle x,y\rangle}{\lambda^2|x||y|} = \angle(x,y)
For the converse suppose \lambda_i \ne \lambda_j for some i,j. Then \angle(Tx_i, T(x_i+x_j)) \ne \angle(x_i, x_i+x_j) (I think, too lazy to check rigorously)
Let x \ne 0, we have
Tx = T(x_1,x_2) = (x_1\cos\theta + x_2\sin\theta, -x_1 \sin\theta + x_2\cos\theta)
And
\begin{aligned} \langle Tx, x\rangle &= x_1^2\cos\theta + x_1x_2\sin\theta - x_1x_2\sin\theta + x_2^2\cos\theta \\ &= (x_1^2 + x_2^2)\cos\theta \end{aligned}
Note T is norm preserving (by some easy algebra and the fact that \cos^2\theta + \sin^2\theta = 1), and so finally
\angle(Tx,x) = \arccos \frac{\langle Tx,x\rangle}{|Tx|\cdot |x|} = \arccos \frac{(x_1^2+x_2)^2\cos\theta}{x_1^2 + x_2^2} = \theta
Let e_1,\dots,e_n be the standard basis, apply the triangle inequality and Cauchy-Swartz to get
\begin{aligned} |T(h)| &= |T(a_ie^i)| \\ &\le \sum_{i=1}^n |a_i| |Te_i| \\ &\le \underbrace{\left(\sum_{i=1}^n |a_i|^2\right)^{1/2}}_{|h|} \cdot \underbrace{\left(\sum_{i=1}^n |Te_i|^2\right)^{1/2}}_M \\ &= |h|\cdot M \end{aligned}
Easy to check mentally via the summations. I particularly like |(x,z)| = \sqrt{|x|^2 + |z|^2}. It’s like you can have “coordinates” where x \in \mathbf{R}^n and z \in \mathbf{R}^m :D
If Tx = \varphi_x = 0 then \langle x,y\rangle = 0 for all y, set y = x to get |x|^2 = 0 implying x = 0. Thus, T is 1-1, and by the Fundamental Theorem of linear algebra T is onto as well. Meaning every \varphi can be written \varphi = Tx for a unique x.
|x+y|^2 = \langle x+y,x+y\rangle = |x|^2 + |y|^2 + 2\langle x,y\rangle
If \langle x,y\rangle = 0 we clearly have |x+y|^2 = |x|^2+|y|^2.
It’s weird that Spivak uses rectangles instead of circles for open sets.