User:TakuyaMurata/Intro to Analysis/Linear space

An function, we call a function such as this "a linear operator", for our desire to avoid sounding awkward by speaking of functions of functions.

We define a differential operator $$\mathit{d}$$, a special form of a linear operator by d(x^n) = n x^(n - 1) where x is a variable and n is an integer. Now, we can differential a polynomial by d and a power series; there are functions, however, that cannot be expressed in a combination, whether finite or infinite, of terms, namely non-analytic function.

When f(x) = x for some element x, possibly set, we say the operator f is identity or invariant on x, or equivalently, x is invariant under f.

The simplest of a basis would be 1 and -1 in the field of positive numbers. It is easy to see that every number n can be written as n = a(1) + b(-1) and the expression is unique.

A linear space is a field with a basis. The dimension is The order of in the basis is said to be

Theorem. A liner space of each finite dimension is unique up to isomorphism. Proof. Since bases are finite and has the same order, there is a bijection between.

For each vector a, a norm of a, denoted by \|z\|, is a complex number with the following properties: (1) $$\|x\| \ne 0 for x \ne 0$$. (2) $$\|z_0 + z_1\| \le \| z_0 \| + \| z_1 \|$$. (triangular inequality)

Some properties follow immediately:

Theorem. Given a vector x,

(1) $$\| x \| \ge 0$$ (2) $$\| \| x \| - \| y \| \| \le \| x - y \|$$. Proof. (1) Do the computation: $$\| x - x \| \ge \| x \| + \| -x \| = 2\| x \|$$. (2) $$\| x \| = \| x - y + y \| \le \| x - y \| + \| y \|$$. Then \| x \| - \| y \| \le \| x - y \|. Swapping symbols x and y leads to what we want.

We say the norm at $$x$$ is continuous if the function f(x) = ||x|| is continuous.

Theorem. $$\|x\|^2 = $$. Proof.

Theorem. ''A norm |x| for a vector x has the following properties: if a is a scalar, then (1) $$\|az\| = |a| ||z||$$. (2) $$\|a\| = |a|$$. (3) $$\|z\| = 0$$ if and only if z is a zero vector. Proof. (1) ||az||^2 =  = a = a \bar  = a \bar a  = |a|^2 ||z||^2. (2) Let ||z|| = 1, then from (1) ||a|| = \|az\| = |a| \|z\| = |a|. (3) If ||z|| = 0, then  = 0. The converse follows from (2).

Theorem. If a linear function f is continuous at the vector 0, then f is a bounded operator; i.e., $$f(x) \le C \| x \|$$ for all x. Proof. Suppose f is not a bounded operator. That is, suppose a sequence x_n converging to x such that $$\| f(x_n - x) \| \ge n \| x_n - x \|$$. Then

Theorem. If f is a bounded operator, then f'' is continuous. Proof.

Combining previous two previous two theorems gives:

Corollary. If a linear function f is continuous at x_0, then f is continuous everywhere.