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Say I have a continuous function $f$ defined on a compact interval $I$ on the real line. As is well-known, I could approximate $f$ arbitrarily well by polynomials.

Given $R>0$, how well can we approximate $f$ as a linear combination of functions of the form $x^r$, where $r$ lies in $\lbrack 0,R\rbrack$? If $f$ is analytic, can we express $f$ as an integral $\int_0^R x^r d\mu(r)$?

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The "i.e." part at the end of the first sentence doesn't follow from the possibility of approximating by polynomials. The approximating polynomials need not be the partial sums of an infinite series. –  Andreas Blass Nov 3 '12 at 13:20
    
Ah, you are right. Well, I was really thinking of $f$ analytic. –  H A Helfgott Nov 3 '12 at 16:49
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As noticed in the previous remark, it is not true that every $f$ can be represented by a power series, $\sum c_kx^k$ because the sum of this series must be analytic. Same applies to the intergal $\int x^r d\mu(r)$. This function is analytic on intervals $I$ that do not contain zero.

However, there is an analog of Weierstrass theorem for non-integer powers $x^{r_k}, r_k>0$. This is called the Muntz-Sasz theorem. It says roughly speaking that the span of these monomials is dense in C[a,b], 0 < a < b, if and only if the series $\sum 1/r_k$ diverges.

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Let us assume that your $f$ is continuous and compactly supported on the real line. Now for $\epsilon >0$, just consider the following convolution by a Gaussian mollifier $$ f_\epsilon(x)=\int e^{-\pi\epsilon^{-2}(x-y)^2} f(y) \epsilon^{-1}dy $$ Obviously the fonction $f_\epsilon$ is entire and also is converging uniformly towards $f$: we have $$ f_\epsilon(x)-f(x)=\int e^{-\pi y^2}\bigl( f(x+\epsilon y)-f(x)\bigr) dy, $$ so that $ \vert f_\epsilon(x)-f(x)\vert\le \int_{\vert y\vert \le \lambda} e^{-\pi y^2}\bigl\vert f(x+\epsilon y)-f(x)\bigr\vert dy + \int_{\vert y\vert \ge \lambda} e^{-\pi y^2}\bigl\vert f(x+\epsilon y)-f(x)\bigr\vert dy $ and $$ \vert f_\epsilon(x)-f(x)\vert\le \sup_{\vert x_1-x_2\vert\le \epsilon \lambda}\vert f(x_1)-f(x_2)\vert + \int_{\vert y\vert \ge \lambda} e^{-\pi y^2} dy 2\Vert f\Vert_{L^\infty(\mathbb R)}. $$ Choosing for instance $\lambda=\epsilon^{-1/2}$, and using uniform continuity of $f$, we prove uniform convergence. Note that it is an easy way to prove Stone-Weierstrass theorem on $\mathbb R$ since the entire function $f_\epsilon$ can be uniformly approximated by polynomials on compact sets.

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Why would $f_\epsilon(x)$ be a linear combination of functions of the form $x^r$? Or are you doing a change of variables? –  H A Helfgott Nov 4 '12 at 0:40
    
$f_\epsilon$ is an entire function, thus can be approximated uniformly by polynomials on compact sets. –  Bazin Nov 9 '12 at 18:59
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