# Is the set of all polynomials in $\log(x)$ dense in $L^2[0,1]$?

Is $\{(\log(x))^k\mid k=0,1,2,\ldots\}$ dense in $L^2 [0,1]$? That is, is the set of all polynomials of logarithm functions dense in the set of square integrable functions on $[0,1]$?

• @Fabian I think the OP is asking wether the linear span of the polynomials is dense in $L^2$. The fact that they all vanish at $0$ (except for $k=0$ btw) is of no consequence since we are working in $L^2$. – Olivier Bégassat Jan 2 '13 at 15:14
• How do you evaluate your $(\log(x))^k$ at $0$? You seem to work on the closed interval $[0,1]$ after all... – mkl Jan 2 '13 at 15:29
• @mkl The space $L^2$ consists of equivalence classes of functions; pointwise evaluation is not defined for elements of $L^2$. – user53153 Jan 3 '13 at 23:33
• @Pavel you're right, of course. I was led somewhat astray by remarks of the functions vanishing at 0, a single point, after all. – mkl Jan 4 '13 at 7:45

Let's change the variable: $$x=e^{-t}$$, $$t> 0$$. This induces an isometry between $$L^2([0,1])$$ and the weighted Lebesgue space $$L^2((0,\infty),e^{-t})$$. The polynomials in $$\log x$$ become polynomials in $$t$$. We may decide to orthonormalize them with respect to the weight $$e^{-t}$$, thus obtaining Laguerre polynomials. The question can now be restated as: do the Laguerre polynomials form a basis of $$L^2((0,\infty),e^{-t})$$? Interestingly, I could not find an answer to this obvious question in the Wikipedia article, or Mathworld article, or in Springer EOM. At last a Google search brought up this paper where this statement is made explicitly:

It is known that $$\{L_n\}$$ is a basis in $$L^2(0,\infty)$$ with respect to the measure $$e^{-x}\,dx$$ (see, e.g., [6, p. 349]).

 G. Sansone, Orthogonal functions, rev. English ed., Dover Publications Inc., New York, 1991.

This is not an answer (and possibly total nonsense), but it's too long for a comment. Let $V=\operatorname{Vect}(1,\log,\log^2,\dots)$, and suppose we wanted to show it is dense (which I suspect). We only need to show that its orthogonal is $0$.

We could try working with a Hilbert basis. I don't know how to calculate the integrals of a cosine/sine-function against $\ln^q$. However, for $p,q\in\mathbb{N}$: $$\langle t^p,\log^q\rangle=\frac{(-1)^q q!}{(p+1)^{q+1}}$$ Because continuous maps are dense in $L^2[0,1]$ and continuous functions can be uniformly approximated using polynomials, polynomials are dense in $L^2$. Suppose $f\perp V$, and $f\neq 0$. There should be real numbers $c_n,~n\in\Bbb N$ with $f=\sum_{n\in \Bbb N} c_n t^n$ in $L^2$ (which strikes me as perfectly unbelievable). If this holds, letting $c_N$ be the first non-zero coefficient, we'd probably get $$0=\langle f,\ln^q\rangle\sim c_N\frac{(-1)^q q!}{(N+1)^{q+1}}$$ which would be contradictory, and so $V$ would be dense.

Any input appreciated!

• Seems like a case for Stone-Weierstrass to show first that the linear span if the logarithms is dense in $C[\varepsilon,1]$, which is enough. – Martin Jan 3 '13 at 15:19
• @Martin I'm not approximating the logarithms, but continuous funtions on $[0,1]$. – Olivier Bégassat Jan 3 '13 at 15:20
• So am I :-)  – Martin Jan 3 '13 at 15:21
• @Martin sorry, I might have misunderstood your comment. Do you mean to say that we can first show that the polynomials in $\ln$ can approximate a continuous funtion on $[\epsilon,1]$? – Olivier Bégassat Jan 3 '13 at 15:22
• Yes: the polynomials in ln contain the constants and they separate the points on $[\varepsilon,1]$ and they are an algebra, so Stone Weierstrass tells us they are uniformly dense in $C[\varepsilon,1]$ (for real scalars). – Martin Jan 3 '13 at 15:24