# The set of all continuous functions $f:[0,1]\rightarrow R$ satisfying $\int_{0}^{1}t^{n}f(t)dt=0$

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? –  David Mitra Dec 9 '12 at 17:36
Don't use dollar symbols $for regular text!! – Nameless Dec 9 '12 at 17:37 The set is nonempty because it contains the zero function – Amr Dec 9 '12 at 17:39 add comment ## 2 Answers Consider the map$\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. - But why is$\mathbb R[x]$dense? That's most of the question. – Michael Hardy Dec 9 '12 at 17:44 @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. – Alex Youcis Dec 9 '12 at 17:45 The statement in question is the Weierstrass Approximation Theorem. The Stone-Weierstrass Theorem is something considerably more general. – Pete L. Clark Dec 10 '12 at 18:44 @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" – Alex Youcis Dec 10 '12 at 19:00 @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... – Pete L. Clark Dec 10 '12 at 19:44 show 2 more comments 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. –  learner Dec 9 '12 at 18:01