# Does this "show" that the fourier basis is more "powerful" than the polynomial basis?

I'm sure the answer is negative, they are both called "basis" after all, but there is a "paradox" that I can't wrap my head around.

We can write the dirac delta function in the fourier basis as $$\delta(x) = \frac{1}{2\pi} + \frac{1}{\pi}\sum_{n=1}^\infty \cos nx$$ yet since the delta function is not a "function" per se, it seems that it can't have a taylor series. Does this mean that it can't be expressed in the polynomial basis?

I guess there are 2 possible answers to this question:

1. It can still be expressed in the polynomial basis, just by coefficients that are different from those given by a taylor series
2. Since the dirac delta is not a function, "basis" theory loses its meaning, and so it does make sense that it can only be expressed by sins and coses. In this sense, is the fourier basis "more powerful/expressive" than the polynomial basis (ie it can also express things such as distributions)
• Any basis can express the space it spans. If you choose a polynomial basis that spans the same space then it will be able to span the same space, but I don't see any specification on the polynomial basis, nor which inner product is used. May 25 '17 at 21:42
• I think Fourier Transform is usually first defined on the quite restrictive Schwarz class of infinitely continously differentiable functions ($\mathbf C^\infty$) with compact support, which is so odd that is usually required to show the students a proof that it's in fact not empty. Then we can find families of functions which we can use to help expand the transform to other spaces of functions. May 25 '17 at 21:56
• @mathreadler Quibble: this is Laurent Schwartz with a "t"... also, "the Schwartz space" of functions is nowadays often the collection of smooth functions so that they and all derivatives are rapidly decreasing, while the space of smooth, compactly-supported functions is (Schwartz') test functions. May 25 '17 at 22:09
• @paulgarrett: yes you are right maybe I risked messing up the wrong Schwar(t)z:es, it is late over here. Is the one I spelled the one with the famous inequality in linear algebra? Yes maybe the space of test functions, but that's what distributions are defined for after all and the question regards the Dirac distribution which works on such test functions from that space so it would only make sense to talk about the most restrictive of the two spaces, right? May 25 '17 at 22:18
• @mathreadler, heh, yes, the linear-algebra inequality-Schwarz is Herman Amandus (or similar) Schwarz. And, yes, there is ambiguity in "Schwartz' space...", though/and the dual to (what I call) Schwartz' space, a.k.a. tempered distributions, is smaller than the dual of the smaller space of (what I call) test functions. But of course these are just descriptors... :) May 25 '17 at 22:21

It's not a Taylor series, but you can also write the Dirac delta as a sum (convergent in the sense of distributions on the interval $(-1,1)$) of Chebyshev polynomials $T_n(x)$:
$$\delta(x) = \frac{1}{\pi} + \frac{2}{\pi} \sum_{k=1}^\infty (-1)^k T_{2k}(x)$$
A related example where things turn out somewhat better is the case of the "quantum harmonic oscillator" $-\Delta+x^2$ on $\mathbb R$, which does have an orthonormal basis of eigenfunctions in $L^2(\mathbb R)$, and whose subtler theory does allow expression of tempered distributions as infinite sums of terms which are constant multiples of $n$th Hermite polynomial times suitable Gaussian.