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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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    Have you seen [this](http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem)?2012-12-09
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    Don't use dollar symbols $ for regular text!!2012-12-09
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    The set is nonempty because it contains the zero function2012-12-09
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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

2 Answers 2

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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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    But why is $\mathbb R[x]$ dense? That's most of the question.2012-12-09
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    @MichaelHardy Presumably anyone given this question would know about the Stone-Weierstrass theorem, so I didn't think that it needed to be said. But, you are probably correct, I've edited.2012-12-09
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    The statement in question is the Weierstrass Approximation Theorem. The Stone-Weierstrass Theorem is something considerably more general.2012-12-10
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    @PeteL.Clark I guess I was using an imprecise version of "precise". I mean to say "it's the part of Stone-Weierstrass Theorem that is relevant"2012-12-10
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    @Alex: right, and it's certainly no big deal, but: the result being used *is* **precisely** the Weierstrass Approximation Theorem. In particular this (very standard) application was surely known to Weierstrass almost 20 years before Stone was born. We may as well give credit where it is actually do...2012-12-10
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    @PeteL.Clark Good point--I even learned something (I didn't know that Weierstrass's (approximation) theorem predated the general Stone-Weierstrass theorem by that long. Edited. Thanks!2012-12-10
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    @Alex:...+1. :)2012-12-10
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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