What is my operator norm (cannot get good enough bounds). Given a space of square integrable functions $x(t)$ over the interval $[0;1]$ one can introduce a norm
$$\|x(t)\|= \sqrt{\int_0^1 (x(t))^2 \, dt};$$
Then what is a norm of the transformation below (result is also square integrable function, same space, same norm) assuming the definition.
$$\operatorname{A}x(t)= \int_0^ts \cdot x(s) \, ds;$$

Well according to one definition $\|\operatorname{A}\| = \sup_{\|x\|=1}\|\operatorname{A}x(t)\|\ge \|\operatorname{A}1\| =\sqrt{\frac{1}{20}}$, I tried quite a few different functions and ain't getting any better lower bound than this. To be specific I tried $x(t) = t^\alpha/\| t^\alpha \|$ (for $\alpha \in (-0.5;+ \infty)$) - result is $\sqrt{\frac{1}{20}}$ or lower.
According to other equivalent definition I look for infimum of bounding constants
$$\|\operatorname{A}x(t)\| = \sqrt{\int_0^1 \left(\int_0^t s \cdot x(s) \, ds \right)^2 \, dt} \le \|x(t) \| \cdot \sqrt{\int_0^1 \int_0^t s^2 \, ds \, dt} = \sqrt{\frac{1}{12}} \cdot \|x(t) \|;$$
As a result $\sqrt{\frac{1}{20}} \le \|x(t) \| \le \sqrt{\frac{1}{12}}$. How can I improve?
 A: This is only a partial answer.
Consider ${\rm A}$ as a Hilbert–Schmidt integral operator, i.e. for every $x\in L^2[0,1]$,
$${\rm A}x(t)=\int_0^1k(t,s)x(s)ds,\quad t\in[0,1],$$
where the kernel $k$ of $A$ is given by:
$$k(t,s)=s\cdot 1_{[0,t]}(s),\quad t,s\in[0,1].$$
As a result,
$$\|{\rm A}\|\le \|k\|_{L^2([0,1]^2)}=\frac{1}{2\sqrt{3}}.$$
Let ${\rm A}^*$ be the adjoint operator ${\rm A}^*$ of ${\rm A}$, i.e. $\langle{\rm A}x,y\rangle=\langle x,{\rm A}^*y\rangle$ for any $x,y\in L^2[0,1]$. By definition,
$${\rm A}^* x(t)=\int_0^1k(s,t)x(s)ds.$$
It follows that for ${\rm B}:={\rm A}^*{\rm A}$,
$${\rm B}x(t)=\int_0^1K(t,s)x(s)ds,\quad t\in[0,1],$$
where the kernel $K$ of $B$ is given by:
$$K(t,s)=\int_0^1k(r,t)k(r,s)dr=(1-\max(t,s))ts,\quad t,s\in[0,1].$$
By definition,
$$\|{\rm A} x\|^2=\langle{\rm B}x,x\rangle.$$
Note that ${\rm B}$ is a compact self-adjoint operator and ${\rm B}$ is positive definite, so from the spectral theorem we know that
$$\|{\rm A}\|^2=\sup_{\|x\|=1}\langle{\rm B}x,x\rangle=\|{\rm B}\|=\text{ maximal eigenvalue of } {\rm B}.$$
Let $\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_i\ge\cdots>0$ be all the all eigenvalues of $B$(counting multiplicity). Then
$$\|{\rm B}^n\|=\lambda_1^n,\quad n\ge 1.$$
Also note that $B^n$ is a Hilbert–Schmidt integral operator with kernel $K_n$, where
$K_1=K$ and 
$$K_{n+1}(t,s)=\int_0^1K_n(t,r)K(r,s)dr,\quad t,s\in[0,1],~ n\ge 1.$$
It follows that
$$\|K_n\|_{L^2([0,1]^2)}={\rm tr}{\rm B}_n=\sum_{i=1}^\infty \lambda_i^n,\quad n\ge 1. $$
This provides a practical way to estimate the upper bound of $\|{\rm A}\|$:
$$\|{\rm A}\|=\|{\rm B}^n\|^{\frac{1}{2n}}\le\|K_n\|_{L^2([0,1]^2)}^{\frac{1}{2n}},\quad n\ge 1,$$
and 
$$\|{\rm A}\|=\lim_{n\to\infty}\|K_n\|_{L^2([0,1]^2)}^{\frac{1}{2n}}.$$
In particular, for $n=1$, we have a rough upper bound
$$\|{\rm A}\|= \|B\|^{\frac{1}{2}}\le\|K\|_{L^2([0,1]^2)}^{\frac{1}{2}}=6^{-\frac{1}{2}}\cdot7^{-\frac{1}{4}}<\frac{1}{2\sqrt{3}}.$$

Another way to estimate $\lambda_1=\|{\rm B}\|=\|{\rm A}\|^2$ is to notice that the eigenspace
$$E_1:=\{x\in L^2[0,1]\mid {\rm B}x=\lambda_1 x\}$$
contains some positive function. It can be seen as follows. If $x\in E_1\setminus\{0\}$, then $x=x^+-x^-$, where $x^\pm:=\max(\pm x,0)$. We may assume that $x^+\ne 0$. Since $y\ge 0\Rightarrow {\rm B}y\ge 0$, 
$$Bx^+\ge \max(Bx,0) =(Bx)^+=\lambda_1x^+\Rightarrow \langle Bx^+,x^+\rangle\ge \lambda_1\langle x^+,x^+\rangle.$$
Recall that $\lambda_1=\sup_{\|y\|=1}\langle{\rm B}y,y\rangle$. Therefore,
$$0<\langle Bx^+,x^+\rangle=\lambda_1\langle x^+,x^+\rangle \Rightarrow x^+\in E_1\setminus\{0\}.$$
Now let $x\in E_1\setminus\{0\}$ with $x\ge 0$. Since  $\langle x,1\rangle>0$, we can write $1$ as 
$$1=x_1+x_2,\quad x_1\in E_1\setminus\{0\},~ x_2\perp E_1.$$
It follows that
$$\lim_{n\to \infty}\lambda_1^{-n}\cdot{\rm B}^n 1=x_1,$$
i.e. we an use the formula 
$$\lambda_1=\lim_{n\to \infty}\|{\rm B}^n 1\|^{\frac{1}{n}}$$
to estimate $\|{\rm A}\|$.
A: Let $\mu_n$ be the $n$-th positive root of the Bessel function of the first kind $J_{-1/4}$. 
Functions $x^{3/2}J_{-1/4}(\mu_n x^2)\,$ are orthogonal for different $n$:
$$
\int_0^1 x^{3} J_{-\frac{1}{4}}(x^2 \mu_m)J_{-\frac{1}{4}}(x^2 \mu_n)\,dx=
\frac12\int_0^1 y J_{-\frac{1}{4}}( \mu_my)J_{-\frac{1}{4}}(\mu_ny)\,dy=
\frac14J_{3/4}^2(\mu_n)\delta_{mn}.
$$
The last equality is formula $(53)$ for $a=1$ from the cited page.
It follows in particular, what $J_{3/4}(\mu_n)\ne0$, $n\in \mathbb N$. So the functions 
$$
j_n(x)=\frac{2x^{3/2}J_{-1/4}(\mu_n x^2)}{J_{3/4}(\mu_n)}
$$
form an orthonormal sequence on $[0,1]$: $$(j_m,j_n)=\delta_{mn}.$$
Using Mma for calculations gives
$$
Aj_n(x)=\int_0^xtj_n(t)\,dt=\frac{x^{3/2}J_{3/4}(\mu_n x^2)}{J_{3/4}(\mu_n)},
$$
$$
(Aj_m,Aj_n)=\frac{\delta_{mn}}{4\mu_n^2}.
$$
Now lets expand a function $f\in L_2([0,1])$ into the series of $j_n$:
$$
f(x)=\sum_{n=0}^{\infty}a_nj_n(x).
$$
Up to the transform $x\rightarrow x^2$ it is the Fourier–Bessel series. 
Using the orthogonality properties of $j_n$ and $Aj_n$ stated above, we have
$$
\frac{(Af,Af)}{(f,f)}=
\frac{\displaystyle\frac14\sum_{n=1}^{\infty}\frac{a_n^2}{\mu_n^2}}{\displaystyle\sum_{n=1}^{\infty}a_n^2}.
$$
Since the sequence $\mu_n$ is increasing, the maximum of the lhs is attained for $f=j_1$ and
$$
\|A\|=\frac1{2\mu_1}=0.249215\ldots
$$
