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The first one:

$$a) A(x(t)) = \int_{0}^{t}x(s)ds$$

Here I proved that $\|A(x(t))\|_{L_2[0,1]} \leq \|x(t)\|_{L_2[0,1]}$. Unfortunately I cannot find a function that has the norm of one and the map having the norm of one too.

$$b) \lambda \in (0,1), A(x(t)) = \begin{cases} x(t), t \leq \lambda \\ 0, t > \lambda \end{cases} t\in [0,1]$$

$\|A(x(t))\|_{L_2[0,1]} \leq \|x(t)\|_{L_2[0,1]}$ I got this again, but I am sure the norm is less than one. Closer to $\lambda$. But I dont know how to prove this.

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    $\begingroup$ Integrate x(s) = s, see what you get then normalize it so that it has norm 1. $\endgroup$
    – Paul
    May 8, 2017 at 13:38
  • $\begingroup$ @Paul what does that hint mean? $\endgroup$ May 8, 2017 at 18:04

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First of all, the proof of the "other direction" is important for motivating an example. Here's my version: $$ \|Ax\| = \int_0^1 \left|\int _0^tx(s)\,ds\right|^2 dt \leq \int_0^1 \int _0^t |x(s)|^2\,ds\, dt \\ = \int_0^1 \int_s^1 |x(s)|^2 \,dt\,ds = \int_0^1\, (1-s)|x(s)|^2\,ds\\ \leq \int_0^1 |x(s)|^2\,ds $$ Now, from that second to last step, perhaps you can intuit that for any $x(s)$, we will have $\|Ax\| < \|x\|$. However, we're ultimately interested in a supremum. Ultimately, what we really want is to find an $x_\epsilon(t)$ such that $\|A x_\epsilon\|/\|x_\epsilon\| \to 1$ as $\epsilon \to 0^+$. Intuitively, such a function should have most of its "weight" on the left side.

In particular, consider $$ x_\epsilon(t) = \begin{cases} 1/\sqrt{\epsilon} & t < \epsilon\\ 0 & t \geq \epsilon \end{cases} $$ Calculate $\|x_\epsilon\|^2 = 1$, whereas $$ Ax_\epsilon(t) = \begin{cases} t/\sqrt{\epsilon} & t < \epsilon\\ \sqrt{\epsilon} & t \geq \epsilon \end{cases} \\ \|Ax_\epsilon\|^2 = (1 - \epsilon)\epsilon + \int_0^\epsilon t^2/\epsilon\,dt = (1 - \epsilon)\epsilon + \frac 13 \epsilon^2 $$

For $b$: in fact, the norm of the operator is $1$. Consider $$ x(t) = \begin{cases} 1 & t \leq \lambda\\ 0 & t \geq \lambda \end{cases} $$ Verify that $Ax = x$, so that $\|Ax\| = \|x\|$. As a linear operator, we might regard this $A$ as an orthogonal projection onto a subspace.

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  • $\begingroup$ You should have $Ax_\epsilon(t) = \sqrt{\epsilon}$ for $t\geq \epsilon$. With that you should get that $||A x_\epsilon||^2 = \frac{\epsilon^2}{3} + (1-\epsilon)\sqrt{\epsilon}$ instead. $\endgroup$ Sep 28, 2023 at 15:00

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