Спасибо. Похоже, что понял принцип. Например если дан оператор
=e^{-t}x(\sqrt{t})$ $A[x](t)=e^{-t}x(\sqrt{t})$](https://dxdy-01.korotkov.co.uk/f/4/4/b/44b246a057e130d56dad13a690bce49a82.png)
в
![$L_2[0,1]$ $L_2[0,1]$](https://dxdy-02.korotkov.co.uk/f/9/9/9/9995872b7fae100e1a05b0003beffe9a82.png)
.
![$$\|x\|\leqslant 1\Rightarrow \|Ax\|=\left(\int\limits_0^1
e^{-2t}|x(\sqrt{t})|^2dt\right)^{1/2}=\left(\int\limits_0^1
2te^{-2t^2}|x(t)|^2dt\right)^{1/2}\leqslant\left(\int\limits_0^1\max\limits_{t\in[0,1]}(2te^{-2t^2})|x(t)|^2dt\right)^{1/2}$$
$$\left(\int\limits_0^1\max\limits_{t\in[0,1]}(2te^{-2t^2})|x(t)|^2dt\right)^{1/2}=2^{1/4}e^{-1/2}\|x\|\leqslant2^{1/4}e^{-1/2}.$$ $$\|x\|\leqslant 1\Rightarrow \|Ax\|=\left(\int\limits_0^1
e^{-2t}|x(\sqrt{t})|^2dt\right)^{1/2}=\left(\int\limits_0^1
2te^{-2t^2}|x(t)|^2dt\right)^{1/2}\leqslant\left(\int\limits_0^1\max\limits_{t\in[0,1]}(2te^{-2t^2})|x(t)|^2dt\right)^{1/2}$$
$$\left(\int\limits_0^1\max\limits_{t\in[0,1]}(2te^{-2t^2})|x(t)|^2dt\right)^{1/2}=2^{1/4}e^{-1/2}\|x\|\leqslant2^{1/4}e^{-1/2}.$$](https://dxdy-04.korotkov.co.uk/f/b/4/0/b40a151ab33c10030198ec8a696c3e9682.png)
Строим функции:
![$$x_\varepsilon(t)=\begin{cases}1,\
t\in(1/\sqrt{2}-\varepsilon,1/\sqrt{2}+\varepsilon)\\0,\
t\in[0,1]\setminus(1/\sqrt{2}-\varepsilon,1/\sqrt{2}+\varepsilon)\end{cases}\Rightarrow\
x_\varepsilon(\sqrt{t})=\begin{cases}1,\
t\in((1/\sqrt{2}-\varepsilon)^2,(1/\sqrt{2}+\varepsilon)^2)\\0,\
t\in[0,1]\setminus((1/\sqrt{2}-\varepsilon)^2,(1/\sqrt{2}+\varepsilon)^2)\end{cases}.$$ $$x_\varepsilon(t)=\begin{cases}1,\
t\in(1/\sqrt{2}-\varepsilon,1/\sqrt{2}+\varepsilon)\\0,\
t\in[0,1]\setminus(1/\sqrt{2}-\varepsilon,1/\sqrt{2}+\varepsilon)\end{cases}\Rightarrow\
x_\varepsilon(\sqrt{t})=\begin{cases}1,\
t\in((1/\sqrt{2}-\varepsilon)^2,(1/\sqrt{2}+\varepsilon)^2)\\0,\
t\in[0,1]\setminus((1/\sqrt{2}-\varepsilon)^2,(1/\sqrt{2}+\varepsilon)^2)\end{cases}.$$](https://dxdy-03.korotkov.co.uk/f/2/6/b/26b1e1d2fb2a5b2787ad6b4111e487f982.png)



.
Думаю, можно вывести формулу для поиска нормы в похожих случаях.
Пусть
=\phi(t)x(p(t))$ $A[x](t)=\phi(t)x(p(t))$](https://dxdy-02.korotkov.co.uk/f/1/7/d/17d446d28a2661478e9ef81a1dbffda482.png)
оператор в
![$L_2[a,b]$ $L_2[a,b]$](https://dxdy-02.korotkov.co.uk/f/5/2/4/524605c117ca275ff7729aa45d88a78182.png)
, где
![$\phi\in C[a,b]$ $\phi\in C[a,b]$](https://dxdy-02.korotkov.co.uk/f/5/a/2/5a2a8a2ea80aea3cf74a80263a8a95e282.png)
, а

- непрерывно дифференцируемая строго монотонная функция, т.ч.
![$p([a,b])=[a,b]$ $p([a,b])=[a,b]$](https://dxdy-01.korotkov.co.uk/f/8/0/a/80a14764588bda1d5879767c69979e9082.png)
.
![$$\|Ax\|^2=\int\limits_a^b
|\phi(t)|^2|x(p(t))|^2dt=\int\limits_a^b|\phi(p^{-1}(t))|^2|x(t)|^2\frac{dp^{-1}}{dt}(t)dt\leqslant\max\limits_{t\in[a,b]}|\phi(p^{-1}(t))|^2\frac{dp^{-1}}{dt}(t)\int\limits_a^b|x(t)|^2dt=$$ $$\|Ax\|^2=\int\limits_a^b
|\phi(t)|^2|x(p(t))|^2dt=\int\limits_a^b|\phi(p^{-1}(t))|^2|x(t)|^2\frac{dp^{-1}}{dt}(t)dt\leqslant\max\limits_{t\in[a,b]}|\phi(p^{-1}(t))|^2\frac{dp^{-1}}{dt}(t)\int\limits_a^b|x(t)|^2dt=$$](https://dxdy-02.korotkov.co.uk/f/d/0/2/d02b7e47fbb23aafaf74ddf8cb01120882.png)
![$$=\max\limits_{t\in[a,b]}\left(|\phi(p^{-1}(t))|^2\frac{dp^{-1}}{dt}(t)\right)\|x\|^2.$$ $$=\max\limits_{t\in[a,b]}\left(|\phi(p^{-1}(t))|^2\frac{dp^{-1}}{dt}(t)\right)\|x\|^2.$$](https://dxdy-02.korotkov.co.uk/f/d/b/8/db83ca767cc8421da1f5fa122659ad7282.png)
![$$|\phi(p^{-1}(t))|^2\frac{dp^{-1}}{dt}(t)\in C[a,b]
\Rightarrow\exists
t_0\in[a,b]:\max\limits_{t\in[a,b]}\left(|\phi(p^{-1}(t))|^2\frac{dp^{-1}}{dt}(t)\right)=|\phi(p^{-1}(t_0))|^2\frac{dp^{-1}}{dt}(t_0).$$ $$|\phi(p^{-1}(t))|^2\frac{dp^{-1}}{dt}(t)\in C[a,b]
\Rightarrow\exists
t_0\in[a,b]:\max\limits_{t\in[a,b]}\left(|\phi(p^{-1}(t))|^2\frac{dp^{-1}}{dt}(t)\right)=|\phi(p^{-1}(t_0))|^2\frac{dp^{-1}}{dt}(t_0).$$](https://dxdy-02.korotkov.co.uk/f/5/2/1/521f376304fdfc63b4a1658a8083980882.png)
![$$x_\varepsilon(t)=\begin{cases}1,\
t\in(t_0-\varepsilon,t_0+\varepsilon)\\0,\
t\in[0,1]\setminus(t_0-\varepsilon,t_0+\varepsilon)\end{cases}.$$ $$x_\varepsilon(t)=\begin{cases}1,\
t\in(t_0-\varepsilon,t_0+\varepsilon)\\0,\
t\in[0,1]\setminus(t_0-\varepsilon,t_0+\varepsilon)\end{cases}.$$](https://dxdy-01.korotkov.co.uk/f/0/d/b/0dbca99db872222e0688f7bf76dce9fa82.png)
Используя интегральную теорему о среднем, получаем.

![$$\|A\|=\left(\max\limits_{t\in[a,b]}\left(|\phi(p^{-1}(t))|^2\frac{dp^{-1}}{dt}(t)\right)\right)^{1/2}$$ $$\|A\|=\left(\max\limits_{t\in[a,b]}\left(|\phi(p^{-1}(t))|^2\frac{dp^{-1}}{dt}(t)\right)\right)^{1/2}$$](https://dxdy-02.korotkov.co.uk/f/d/4/b/d4b2191f53afa68e6116584bfb5a714b82.png)
.