I know that the formula $$(T u)(t) := u( \sqrt{t}),\qquad 0 \le t \le 1,$$ defines a continuous linear operator $T : L^1[0, 1] \to L^1[0, 1]$, and I want to determine its norm.
It's easy to see that $\|T\|\le 2$ but I could'n go further, neither by finding an element which norm is = 2, neither by lowering the upper bound.
We have $$\|Tu\| = \int_0^1 |Tu(t)| \, \mathrm{d}t = \int_0^1 |u(\sqrt{t})| \, \mathrm{d}t = \int_0^1 |u(s)| 2s \, \mathrm{d}s \leq 2\|u\|.$$ To maximize this, you need a function that has support only close to $1$. So let $u$ be the characteristic function of $[1-\frac{1}{n},1]$. Then $$\frac{\|Tu\|}{\|u\|} = n\int_{1-\frac{1}{n}}^1 2s \, \mathrm{d}s = n\left(1-\left(1-\frac{1}{n}\right)^2\right) = 2 - \frac{1}{n}.$$ Letting $n \to \infty$ gives $\|T\| = 2$.