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I was thinking about the problem:

The set of all continuous functions $f:[0,1]\rightarrow R$ satisfying $\int_0^1 t^n f(t) \, dt=0,\qquad n=1,2,\ldots$

(a) $\text{is empty},$ (b) $\text{contains a single element},$ (c) $\text{is countably infinite},$ (d) $\text{is uncountably infinite}.$

I am stuck on it and do not know how to proceed.Please help.Thanks in advance for your time.

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    Yes, $f(t)$ has to be identically $0$.. Because you can approximate it with a polynomial by Stone weierstrauss2016-07-29

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Consider the map $\varphi:C[0,1]\to\mathbb{R}$ given by

$g\mapsto \int_0^1 g(t)f(t)\; dt$

It's easy to see that this map is a continuous linear functional when $C[0,1]$ is given the $\|\cdot\|_\infty$ norm. Note then that by applying the hypothesis you can show that $\mathbb{R}[x]\subseteq\ker\varphi$, but since $\mathbb{R}[x]$ is dense in $C[0,1]$ this implies that $\varphi=0$. Take $g=f$ then to conclude that $f=0$.

EDIT: As Michael Hardy points out, I should probably mention that the density of $\mathbb{R}[x]$ in $(C[0,1],\|\cdot\|_\infty)$ is precisely the statement of the Stone-Weierstrass theorem.

EDIT EDIT: As Pete. L Clark points out, we don't need the complexity of the Stone-Weierstrass theorem (which deals with $C(X)$ for $X$ l.c. Hausdorff) we really only need the special case of $[a,b]$ and the subalgebra being polynomials--this is the Weierstrass Approximation Theorem. This particular case is much easier than the general Stone-Weierstrass theorem. There is a constructive proof using Bernstein polynomials--a Numerical Analysis approximant that has the advantage of approximating a huge class of functions (at least the continuous ones) but with the downside of the rate of convergence being sublinear.

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    The hypothesis is that $\int_0^1t^nf(t)\;dt$ for $n\geq 1$ rather than $n\geq 0$, so it's not immediately clear why all polynomials are in $\ker(\phi)$ rather than just those with zero constant term.2017-06-26
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If $f \in C([0,1] \to \mathbb{R})$, then by the Weirstrass approximation theorem, there's a sequence $\{p_n\}$ of polynomials which tend to $f$ uniformly. If $f$ additionally satisfies the above, we have from linearity of the integral that $0 = \lim_{n\to\infty} \int_0^1 p_n f = \int_0^1 f^2$. Since $f^2$ is a continuous nonnegative function, $f = 0$ everywhere. This shows that there's at most one such function.

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    thank you sir for the clarification.So option (b) would be the right choice,i guess.2012-12-09