Proving that the primitives of the Laguerre functions are uniformly bounded Let $(L_k)_{k\geq 0}$ be the Laguerre polynomials. These polynmials are orthogonal with respect to the inner product: $$\langle f,g\rangle = \int_0^\infty f(x)g(x)\mathrm e^{-x}\,\mathrm dx.$$
Hence, the functions $\psi_k(x) = \sqrt{2} L_k(2x) \mathrm{e}^{-x}$ form a basis of $\mathrm L^2(\mathbb R_+)$ called the Laguerre basis.
I am interested in the functions $\Psi_k(x) = \int_0^x \psi_k(t)\,\mathrm dt$. I can see numerically that these functions are uniformly bounded by $\sqrt 2$ (actually, it seems that they are bounded by $\sqrt 2(1-\mathrm e^{-x})=\Psi_0(x)$, but I just need the constant), but I fail to prove it.
Does anybody know if these functions have already been studied by someone? And if not, does anybody have an idea on how to prove it?

Since the closed form of the Laguerre polynomials is:
$$ L_k(x) = \sum_{j=0}^k \binom{k}{j}\frac{(-x)^j}{j!}, $$
I can show that $\Psi_k$ tends to $(-1)^k\sqrt{2}$ when $x$ goes to $+\infty$, so I need to prove that it's the maximum/minimum of $\Psi_k$.
I tried to use the recurrence relations satisfied by the Laguerre polynomials. For instance, $(\Psi_k)$ satisfy the relation:
$$ \Psi_{k+1}(x) = \frac{2x\,\psi_k(x) - \Psi_k(x) + k \Psi_{k-1}(x)}{k+1}. $$
They also satisfy the relation $\Psi_{k+1}+\Psi_k = \psi_k - \psi_{k+1}$. I tried to use both relations, and nothing came out, but maybe it is the right way.
 A: OK, here is the argument. We want to show that the partial integrals of $u(x)=L_n(x)e^{-x/2}$ are not (much) larger in the absolute value than the full integral.
We'll just use the differential equation
$$
xL_n''+(1-x)L_n+nL_n=0\,.
$$
Plugging $L_n=e^{x/2}u$, we get
$$
x(u''+u'+\tfrac 14 u)+(1-x)(u'+\tfrac 12u)+nu=xu''+u'+(n+\tfrac 12-\tfrac 14x)u=0
$$
Now we want to kill the first derivative, so we define $v(x)=u(x^{3/2})x^{1/2}$ (so that partial integrals of $v$ are the same as of $u$ up to the constant $2/3$ factor). We have
$$
v'(x)=u'(x^{3/2})\tfrac 32x +u(x^{3/2})\tfrac 12 x^{-1/2}
$$
and
$$
v''=u''(x^{3/2})\tfrac 94 x^{3/2}+\tfrac 94 u'(x^{3/2})-u(x^{3/2})\tfrac 14 x^{-3/2}
$$
so, by the differential equation for $u$,
$$
v''=-[\tfrac 94(n+\tfrac 12-\tfrac 14x^{3/2})+\tfrac 14 x^{-3/2}]u(x^{3/2})
$$
and, writing $u(x^{3/2})=\frac{v(x)}{x^{1/2}}$, we finally get
$$
v''+\Phi v=0
$$
where
$$
\Phi(x)=\tfrac 94\left(\tfrac{n+\frac 12}{x^{1/2}}-\tfrac 14x\right)+\tfrac 14 x^{-2}
$$
We don't really care much what this $\Phi(x)$ is. All that will matter is that it is a decreasing convex function that starts positive at $0$ and ends up negative near $+\infty$.
Since $\Phi$ is decreasing, each next hump of $v$ between the zeroes is larger in area than the previous one (the rightmost hump stretches to $+\infty$), so the maximal absolute value of the partial integrals is attained either when we take the full integral (which is fine for us), or if we stop after the penultimate hump. Thus we just need to show that the second option is dominated by the first one.
If the humps have the areas $A_0<A_1<\dots<A_n$ (so that the full integral is $I=A_0-A_1+\dots\pm A_n$), then the sum up to the penultimate hump is $I\mp A_n$. So, we need to show that $A_n\le C|I|$ with some absolute $C>0$. It suffices to show that $A_{n-1}\le cA_{n}$ with some absolute $c<1$ because then $|I|\ge (1-c)A_n$ and $|I\mp A_n|=|A_n-|I||\le \frac c{1-c}|I|$.
Let now $z$ be the last zero of $v$. Assume for definiteness that the last hump is positive. Let $\Psi(x)=\Phi(z)-\lambda(x-z)$ be the unique linear function such that the solution $w$ of $w''+\Psi w=0$ shot up from $x=z,w(x)=0$ stays positive on $[z,+\infty)$ and tends to $0$ at $+\infty$, so $w$ is just the Airy function, properly shifted and stretched. Let us make $w'(z)$ just a tiny bit less than $v'(z)$.
Since the Wronskian $W(v,w)=\det\begin{bmatrix} v & w\\v' & w'\end{bmatrix}$ vanishes at $z$ and $+\infty$ and $W'=(\Phi-\Psi)vw$, we must have $\int_z^{\infty}(\Phi-\Psi)vw=0$, so $\Phi-\Psi$ changes sign. Since $\Phi$ is convex and $\Psi$ is linear, it is possible only if first $\Phi<\Psi$ and later $\Phi>\Psi$, so $W(v,w)<0$ all the way between $0$ and $+\infty$. By our choice of derivative at $z$, $w$ stays below $v$ near $z$. If it ever breaks up, we should have $v=w>0, v'\le w'$ at the first breaking point, so $W(v,w)=vw'-wv'\ge 0$, which has been ruled out. Thus the area of the hump of the rescaled Airy function $w$ with $w'(z)=v'(z)$ is smaller than $A_n$. Since $\Phi>\Psi$ to the left of $z$, we conclude that $A_{n-1}$ is less than the adjacent hump of $w$. Thus $A_{n-1}/A_n\le c$ where $c$ is the ratio of the areas of the penultimate and the ultimate humps of the Airy function. It is a fixed constant less than $1$, so we are done.
In reality, $c\approx 0.365\dots<\frac 12$, so $\frac c{1-c}<1$ and the maximal integral is the full one but, as I said, I currently have no proof of it.
Edit: Details requested by the OP.
I'm using the following comparison principle. Suppose that $\Phi>\Psi$ and we shoot two solutions $v,w$ of the equations $v''+\Phi v=0$ and $w''+\Psi w=0$ from the same point $a$ with the initial data $v(a)=w(a)=0, v'(a)=w'(a)>0$. Then we have $v\le w$ as long as both $v$ and $w$ stay positive. In particular, if both $v$ and $w$ make positive humps, the hump of $v$ is contained in that of $w$, so its area is not larger. The proof is the same Wronskian story as I wrote for the comparison of the infinite humps, only simpler. Assume first that $v'(a)$ is just a tiny bit below $w'(a)$. Then $v$ starts lower and can break through $w$ at some point $b$ with $v(b)=w(b)>0$, $v|_{(a,b]},w|_{(a,b]}>0$ only if $v'(b)\ge w'(b)$. But then we have $W(v,w)(a)=0$, $W(v,w)\le 0$ while $W'(v,w)=(\Phi-\Psi)vw>0$ on $(a,b)$, which is a clear contradiction.
This immediately answers your question about the humps of $v$ and $w$ to the left of the last zero of $v$ (only there we shoot from the right endpoint). As to two adjacent humps, just reflect the left one about the common zero and  use the decreasing property of $\Phi$ to compare (if $a$ is the common zero of two adjacent humps, then you just need the inequality $\Phi(a-t)\ge \Phi(a+t)$ for $t>0$).
A: There is a simple proof of your assertion that $\Psi_k(x)$ tends to $(-1)^k\sqrt{2}$. One has
$$
\Psi_k(x)=\sqrt{2}\int_0^xL_k(2t)e^{-t}dt=\sqrt{2}\sum_{j=0}^k{k \choose j}\frac{1}{j!}(-2)^j\int_0^x t^je^{-t}dt.
$$
Therefore, if you take the limit $x\rightarrow\infty$, you will recognize the $\Gamma$ function that will yield $j!$. What remains can be summed up by the binomial theorem and you will easily get
$$
\lim_{x\rightarrow\infty}\Psi_k(x)=(-1)^k\sqrt{2}.
$$
