Weierstrass Approximation

Let

f \in C [a, b]

, then there is a sequence

(p_{n}) : ℕ_{1} \to ℝ [x]

that converges uniformly to

f

[a, b]

Integrals Equal Zero Implies Function is Zero

Suppose that

f \in C [0, 1]

such that

\int_{0}^{1} f (x) x^{n} d x = 0

for every

n \in ℕ_{1}

then

f = 0

By wat, we obtain a sequence of poly's $(p_{k})$ that converge uniformly o $f$ .

We know that for any $k$ we see that $p_{k} (x) = \sum_{i = 1}^{l_{k}} c_{i} x^{i}$ and therefore

\int_{0}^{1} f (x) p_{k} (x) d x = \sum_{i = 1}^{l_{k}} c_{i} \int_{0}^{1} f (x) x^{n} = 0

Since products maintain uniform convergence, we also know that $f (x) p_{n} (x) \overset{unif}{\to} f {(x)}^{2}$ . And define $H_{n} : [0, 1] \to ℝ$ as $H_{n} (x) : = \int_{0}^{x} f (t) p_{n} (t) d t$ therefore $H_{n} \overset{unif}{\to} H$ where $H (x) = \int_{c}^{x} f^{2} (x) d x$ specifically when $x = 1$ we have that $\int_{0}^{1} f (x) p_{n} (x) d x (= 0) \overset{unif}{\to} \int_{0}^{1} f^{2} (x) d x$ thus $\int_{0}^{1} f^{2} (x) d x = 0$ so that $f = 0$ as needed.

Continuous Extension on Compact Subset

Let

N \in ℝ

and let

K \subseteq [- N, N]

be compact, then show that every continuous function

f

X

may be extended to a continuous function

g

[- N, N]

such that

{‖ f ‖}_{\infty} = {‖ g ‖}_{\infty}

, conclude that every continuous function on

X

is the uniform limit of polynomials.

We construct the new function $g$ as follows, given $p \in X$ then we define $g (p) = f (p)$ , otherwise $p \notin X$ , note that since $X$ is compact it's closed so that $X^{C}$ is open, since $p \in X^{C}$ then we know that there exists some $ϵ \in ℝ^{+}$ such that $B (x, ϵ) \subseteq X^{C}$

Now define $L = {x \in X : x < p}$ and $R = {x \in X : x > p}$ . If both $L = R = \emptyset$ then $X = {p}$ and so since we assumed $p \notin X$ that would be a contradiction, so instead one of these must be non-empty, assume that $L \neq \emptyset$ , we claim that $L = X \cap (- \infty, p - ϵ]$ , this is the case because for any element $s \in B (p, ϵ)$ it's true that $s \notin L$ , but also we do know that $p - ϵ \in X$ , therefore $L$ is the intersection of two closed sets and therefore closed.

Since $L$ is closed then $\sup (L) \in L$ , if it's the case that $R = \emptyset$ then we linearly interpolate from $f (\sup (L))$ to $0$ over $[\sup (L), N]$ , otherwise $R \neq \emptyset$ and we linearly interpolate from $f (\sup (L))$ to $f (\inf (R))$ over $x \in [\sup (L), \inf (R)]$ . Note that by doing this it maintains the sup norm.

The function is continuous as it's a piecewise continuous and at each connection point they are continouous as the left and right limits are equal. Thus $g$ is a continuous extension of $f$ on $[- N, N]$ , due to this we can apply WAT to $g$ to obtain a sequence of polynomials whose limit is $g$ , then by restriction the domain of the polynomails to $X$ then these "new" ones go to $f$ .