In a recent question, I proved that the Fourier-Legendre expansion of the function $f(x)=\text{sign}(x)$ over $(-1,1)$ is given by: $$2\sum_{m\geq 0}\frac{4m+3}{4m+4}\cdot\frac{(-1)^m}{4^m}\binom{2m}{m}\cdot P_{2m+1}(x)\tag{1}$$ with $P_n(x)$ being the $n$-th Legendre polynomial. On the other hand, it is well known that the Fourier series of $f(x)$ over $(-1,1)$ is given by: $$ \frac{4}{\pi}\sum_{m\geq 0}\frac{\sin((2m+1)\pi x)}{(2m+1)}.\tag{2}$$ When considering truncated series, both representations show Gibbs' phenomenon:

enter image description here

This is what happens over the interval $[0,1]$ when we truncate $(1)$ and $(2)$ at $m=50$ (yellow and blue lines, respectively). The red line represents the constant $1$, the green line the constant $\frac{2}{\pi}\text{Si}(\pi)$.

Assuming that $G=\{g_n(x)\}_{n\geq 1}$ is an orthonormal base of smooth functions, spanning the odd functions in $L^2(-1,1)$ (with respect to the usual inner product) we may define the intensity of the Gibbs phenomenon as:

$$ I_G = \limsup_{M\to +\infty}\sup_{x\in(0,1)}\sum_{m=1}^{M}g_m(x)\int_{0}^{1}g_m(y)\,dy. \tag{3}$$

Now, it is not difficult to differentiate $\frac{4}{\pi}\sum_{m=0}^{M}\frac{\sin((2m+1)\pi x)}{2m+1}$, find the explicit locations of its stationary points and prove through a Riemann-sum argument that the intensity of the Gibbs phenomenon for the Fourier base is exactly $\frac{2}{\pi}\int_{0}^{\pi}\frac{\sin x}{x}\,dx = \frac{2}{\pi}\text{Si}(\pi).$ However, the same approach does not work (at least, at first sight) with $(1)$, so:

How to compute the intensity of the Gibbs phenomenon for the Legendre base, i.e. $$ \limsup_{M\to +\infty}\sup_{x\in (0,1)}2\sum_{m\geq 0}\frac{4m+3}{4m+4}\cdot\frac{(-1)^m}{4^m}\binom{2m}{m}\cdot P_{2m+1}(x) $$ ?

As a second point,

How $I_G$ depends on the chosen base $G$ of $L^2(-1,1)$? What if we change the inner product?

  • 1
    $\begingroup$ It may be useful to know that the following approximation holds for the "Legendre kernel": $$\sum_{m=0}^{M}P_m(x)\longrightarrow\frac{1}{\sqrt{2-2x}}-1.$$ $\endgroup$ Aug 29, 2015 at 23:54

1 Answer 1


Addressing the question about the Legendre base: the intensity is the same as for the Fourier base.

This can be seen using the following equiconvergence theorem.

Let $f:[-1,1]\to\mathbb{R}$ be Lebesgue measurable, with $\int_{-1}^1(1-x^2)^{-1/4}|f(x)|\,dx<\infty$. Let $\ell_n(x)=\sum_{k=0}^n a_k P_k(x)$ be a partial sum of the expansion $f(x)\sim\sum_{k=0}^\infty a_k P_k(x)$, and $\phi_n(\cos\theta)=\sum_{k=0}^n b_k\cos k\theta$ be the $n$-th partial sum of the (cosine) Fourier expansion $$\phi(\theta):=f(\cos\theta)\sqrt{\sin\theta}\sim\sum_{k=0}^\infty b_k\cos k\theta.$$ Then $\ell_n(x)-(1-x^2)^{-1/4}\phi_n(x)\underset{n\to\infty}{\longrightarrow}0$ $\color{red}{\text{uniformly}}$ in $|x|<r$ for any fixed $r<1$.

This is the case $\alpha=\beta=0$ of Theorem $9.1.2$ in G. Szegő Orthogonal Polynomials, which states a similar equiconvergence property for expansions in series of the Jacobi polynomials $P_n^{(\alpha,\beta)}(x)$.

Applied to $f(x)=\operatorname{sgn}x$, the behavior of $\ell_n(x)$ around $x=0$ resembles the behavior of the Fourier series of $\phi(\theta)$ around $\theta=\pi/2$, which is $\operatorname{sgn}(\pi/2-\theta)$ "plus something continuous".


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .