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

However, 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. Any help is appreciated.

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    $\begingroup$ Do not deface your questions. To prevent further defacements, I have temporarily locked the question. $\endgroup$ May 9, 2017 at 19:20
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    $\begingroup$ Locked again. Next time we'll temporarily "lock" your account. $\endgroup$
    – user642796
    May 18, 2017 at 9:16

1 Answer 1

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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 $$ \|A\|\ge \|x_n(\sqrt t)\|_1 = \int_0^1n\cdot \chi_{[(1-1/n)^2,1]}(t)\, dt \\= n[1-(1-1/n)^2] = 2 - \frac 1 n $$

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