-2
$\begingroup$

$(a)$ Show that every continuous function $f$ on $[a,b]$ is the uniform limit of polynomials of the form $p_n(x^3)$.

$(b)$ Describe the subspace of $C[-1,1]$ functions which are uniform limits of polynomials of the form $p_n(x^2)$.

Could you please give me hint to solve this problem?

  • 1
    Hint: Use the Muntz-Szasz Theorem – 2017-01-29
  • 0
    what is your conjecture for part $b$? – 2017-01-29
  • 1
    @mathworker21 That's like shooting at a fly with an AK-47. – 2017-01-30
  • 1
    @zhw. And that's exactly what I like to do, utilizing my second amendment. – 2017-01-30

2 Answers 2

0

For part $a$ it is clear that the polynomials of the form $P(x^3)$ form a subalgebra of $C[a,b]$. It clearly contains all of the constant functions and it separates points (because for example the polynomial $x^3$ is injective). So by the Stone-Weierstrass theorem it is a dense subspace of $C([a,b])$

For part $b$ we obtain the even functions. Clearly if a sequence of even functions in $[-1,1]$ converges uniformly then it converges to an even function. It is also clear that if a sequence of polynomials converges to an even function then the sequence consisting on the same polynomials after removing the odd terms also converges uniformly to that function.

  • 0
    Thanks. Is it possible to use Weierstrass Approximation theorem to solve part a? – 2017-01-29
  • 0
    maybe but I don't see how to be honest. – 2017-01-29
  • 0
    Why is your last sentence clear? – 2017-01-30
  • 0
    because the part consisting of the odd terms is an odd funtion. You use the triangle inequality with $P(x)$ and $P(-x)$. @zhw. – 2017-01-30
0

a) Using just Weierstrass: Because $f$ is continuous on $[a,b],$ $f(x^{1/3})$ is continuous on $[a^3,b^3].$ Hence there are polynomials $p_n(x) \to f(x^{1/3})$ uniformly on $[a^3,b^3].$ This implies $p_n(x^3) \to f(x)$ uniformly on $[a,b].$

b) $f$ is even on $[-1,1]$ iff $f$ is the uniform limit of functions of the form $p_n(x^2),$ where the $p_n$ are polynomials. Proof: $\implies:$ Suppose $f$ is even on $[-1,1].$ Then $f$ is the uniform limit of polynomials $q_n(x)$ on this interval. Because $f$ is even, $f(x) = (f(x)+f(-x))/2$ on $[-1,1].$ It follows that $f(x)$ is the uniform limit of $(q_n(x)+q_n(-x))/2$ on $[-1,1].$ Since each $(q_n(x)+q_n(-x))/2$ has the form $p_n(x^2),$ where $p_n$ is a polynomial, we're done. $\impliedby:$ This is the easy direction.