# Find the norm of the operator $Ax(t)=x(\sqrt{t})$

I figured out that it is well-defined and got that $Ax(t)=x(\sqrt{t})$ and $A:L_1[0,1] \to L_1[0,1]$

But I cannot find a function where this is the norm, so I assume that the norm is less than one. I need to prove this, and find a function to which this applies. All input is well received.

• Do not deface your questions. To prevent further defacements, I have temporarily locked the question. – Daniel Fischer May 9 '17 at 19:20
• Locked again. Next time we'll temporarily "lock" your account. – user642796 May 18 '17 at 9:16

The operator $$A$$ has norm 2. To prove this, observe that $$\|Ax(t)\|_1 = \int_0^1 |x(\sqrt t)|dt = \int_0^1 |x(u)|\cdot 2u\, du\\\le 2 \int_0^1 |x(u)|du = 2\|x(t)\|$$ and, taken the functions $$x_n(t) = n \cdot \chi_{[1-1/n,1]}(t)$$, we get $$\|x_n(t)\|_1=1$$ for all $$n$$, and $$x_n(\sqrt t) = n\cdot \chi_{[(1-1/n)^2,1]}(t)$$, so $$\|A\|\ge \|x_n(\sqrt t)\|_1 = \int_0^1n\cdot \chi_{[(1-1/n)^2,1]}(t)\, dt \\= n[1-(1-1/n)^2] = 2 - \frac 1 n$$