From [a,b] to [$0$,$1$]

Question

I was checking an exercise and it's proof and I couldn't understand why the interval [a,b] can be 'replaced' by [0,1].

This is the exercise:

Prove that every continuous function f:[a,b]$\rightarrow \mathbb R$ is the uniform limit of a sequence of even polynomials if and only if (a,b) doesn't contain the origin.

Proof:

Suppose (a,b) doesn't contain the origin. And for simplicity consider $[0,1]$ instead of [a,b].

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. (and the proof continue)

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So why is it equivalent to work with $[0,1]$ ?

Is there a proof to show that $[0,1] \Leftarrow \Rightarrow [a,b]$ ?

Answer 1

Suppose we have the result for $[0,1].$ If now $f\in C([0,b]),$ define $g:[0,1] \to [0,b]$ by setting $g(x) = bx.$ Then $f\circ g \in C([0,1]).$ So there exists a sequence $P_n$ of even polynomials such that $ P_n \to f\circ g$ uniformly on $[0,1].$ This implies $P_n\circ g^{-1} \to f$ uniformly on $[0,b].$ Since each $P_n\circ g^{-1}$ is an even polynomial, we have the result for $[0,b].$

If now $0<a<b$ and $f\in C([a,b]),$ we can extend $f$ to a continuous function $F$ on $[0,b],$ simply by setting $F(x) = f(a)$ for $x\in [0,a).$ By the above, there is a sequence $P_n$ of even polynomials such that $ P_n \to F$ uniformly on $[0,b].$ This of course implies $ P_n \to F$ uniformly on $[a,b].$ Since $F=f$ on $[a,b],$ we have the desired result for $[a,b].$

The result for intervals $[a,b]$ with $a<b\le 0$ follows easily from this.

Answer 2

Because $f(x) := a+(b-a)x$ is a continuous bijection from $[0,1]$ to $[a,b]$.

Answer 3

Yes, actually you can map $[a,b]$ to $[0,1]$ using the following function: $x \mapsto \frac{x - a}{b - a}$. This is a bijection between the two intervals, and it is also a continuous function as you can check.

Answer 4

This can be proved in two stages:

• Prove $[0,1]\sim [a,1+a]$ (translation)

• Prove $[0,1]\sim [0,a]$ (scale)

Combining these gives the result.

Answer 5

It should be "intuitively obvious" That $0$, $1$ and the distance $|1|$ are arbitrary and can be replaced with $a,b$ so that $b > a$ and $|b -a| > 0$.

More formally $g:(a,b) \rightarrow (0, 1)$ so that $g(a) =0$ and $g(b)$ so that for $x =$ some point a proportion, $p$, of the way between $a$ and $b$ and $g(x)$ is the point the same proportion of the way between $0$ and $1$ is a bijection.

To actually nail the jello to the wall: The proportion $p$ is $\frac {x-a}{b-a}$ so the function $g(x) = \frac {x-a}{b-a}$ is a bijection between $[a,b]$ and $[0,1]$ with $g^{-1}(y) = y*(b-a) + a$. (For $a \le x \le b$ then $0 \le x-a \le b-a$ and $0 \le \frac {x-a}{b-a} \le 1$ so $g:[a,b]\rightarrow [0,1]$. And if $0 \le k \le 1$ then $x = k*(b-a) + a \in [a,b]$ and $g(x) = \frac {(k*(b-a) + a) - a}{b-a} = k$ so $g$ is surjective. And if $a\le x < y\le b$ then $g(x) = \frac {x-a}{b-a} < \frac {y-a}{b-a} = g(y)$ so $g$ is injective.)

..... so in this case $f(x)$ is continuous on $(a,b)$. Then $h(y) = f(y(b-a) + a)=f(g^{-1}(x)) = f(x)$ is continuous on $(0,1)$.