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Let $\mathscr{A}$ be a vector space in $\mathscr{C}[0,1]$ generated by the functions $1,\sin t,\sin ^2t,\cdots.$ Show that $\mathscr{A}$ is dense in $\mathscr{C}[0,1]$.

Theorem $($ Stone Weierstrass Theorem$)$ Let $\mathscr{A}$ be an algebra in $\mathscr{C}(S)$ such that $\mathscr{A}$ contains the constant function $1$ and separates the points of $S$. Then $\mathscr{A}$ is dense in $\mathscr{C}(S).$

A subset $\mathscr{B}\subset \mathscr{C}({S})$ separates the points of $S$ if for any $t,s\in S,\ t\neq s,\ \exists\ f\in \mathscr{B}$ such that $f(t)\neq f(s).$

Now we want to show that $\mathscr{A}$ separates the points of the set $[0,1].$

Can anyone help me please, to show the above conclusion?

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  • $\begingroup$ Hint: given some $f\in C[0,1]$, apply Weierstrass approximation theorem to $g(t)=f(\arcsin t)$. Done. $\endgroup$ – Jack D'Aurizio Nov 6 '17 at 13:15
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Since $\mathscr{A}$ is an algebra

  1. Since, $\mathscr{A} $ is a vector space, for any $f,g\in \mathscr{A},\ \alpha\in \mathbb{R},\ f+g\in \mathscr{A}$ and $\alpha f\in \mathscr{A}$.
  2. Take $f,g\in \mathscr{A}\implies f=\sum_{k=0}^na_k\sin^k t,\ g=\sum_{k=0}^mb_k\sin^k t$. Now, $$fg=\sum_{k=0}^{m+n}c_k\sin^k t\in \mathscr{A}.$$

    so $\mathscr{A}$ is an algebra.

As $\sin$ is an increasing function in the interval $[0,\pi/2]$ and $\pi/2>1.$ If $t,s\in [0,1]$ then $\sin(t)\neq\sin(s).$

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  • $\begingroup$ Isn't $\pi/2>1$? $\endgroup$ – kimchi lover Nov 6 '17 at 3:02
  • $\begingroup$ $\operatorname{span}\{x\}$ is not an algebra. $\endgroup$ – amsmath Nov 6 '17 at 3:03
  • $\begingroup$ @amsmath Yeah, it is not but here it is. Okay, I will elaborate my answer a bit. $\endgroup$ – Sachchidanand Prasad Nov 6 '17 at 3:07
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We can do this with just the Weierstrass theorem: Let $f\in C([0,1]).$ Then $f(\arcsin x) \in C([0,\sin 1]).$ Hence there is a sequence of polynomials $p_n(x) \to f(\arcsin x)$ uniformly on $[0,\sin 1].$ It follows that $p_n(\sin t)\to f(\arcsin (\sin t)) = f(t)$ on $[0,1].$ Since the set of polynomials in $\sin t$ is precisely the given vector space, we're done.

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