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:
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?