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Given is that $T:C([0,a])\rightarrow C([0,a]),\space (Ty)(x)=\frac{x^2}{2}+\int_{0}^{x}ty(t)dt, \space||y||_1= sup_{x \in[0,a]} |y(x)|$. I want to prove that $||Ty-Tz||_1\leq \frac{a^2}{2}||y-z||_1$. My attempt at solving this:
$$sup_{x \in[0,a]} |(Ty)(x)-(Tz)(x)|=sup_{x \in[0,a]}|\int_0^xt(y(t)-z(t))dt|$$ $$\leq sup_{x\in[0,a]}\int_0^x|t(y(t)-z(t))|$$ This is the point at which I made a step that I'm not sure is correct. I got rid of the supremum by replacing the upper bound of the integral with $a$. That is $$\int_0^a|t(y(t)-z(t))|$$ I don't know if this is the correct step to take, but even if it is, I don't know how to proceed. Any help would be appreciated. In particular, explaining how to get rid of the supremum, as that seems to be my main problem here. Thank you in advance.

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You are almost there, it only remains to estimate $|y(t) - z(t)|$ using the norm of the difference: $$ |(Ty)(x) - (Tz)(x)| = \left | \int_0^x t(y(t) - z(t)) \, dt \right | \le \int_0^x t |(y(t) - z(t) | \, dt \\ \le \Vert y - z \Vert_1 \int_0^xt \, dt \, = \Vert y - z \Vert_1 \frac{x^2}{2} \le \Vert y - z \Vert_1 \frac{a^2}{2} \, . $$

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    $\begingroup$ Oh, that's great, I actually had this worked out, I just wasn't sure if the step where you introduce the norm was a valid one to make. I'm glad to see it is. Thanks for answering. $\endgroup$ – user569579 Oct 1 '18 at 17:03
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Note that the integrand

$\vert t (y(t) - z(t)) \vert \tag 1$

of

$\displaystyle \int_0^x \vert t(y(t) - z(t)) \vert \; dt \tag 2$

is non-negative for all $t \in [0, a]$:

$\vert t (y(t) - z(t)) \vert \ge 0, \; t \in [0, a]; \tag 3$

it then follows that, for $x \in [0, a]$, that

$\displaystyle \int_0^x \vert t (y(t) - z(t)) \vert \; dt \le \int_0^a \vert t(y(t) - z(t)) \vert \; dt; \tag 4$

more formally, this may be seen from the equation

$\displaystyle \int_0^x \vert t (y(t) - z(t)) \vert \; dt + \int_x^a \vert t (y(t) - z(t)) \vert \; dt = \int_0^a \vert t(y(t) - z(t)) \vert \; dt, \tag 5$

since (3) implies

$\displaystyle \int_x^a \vert t (y(t) - z(t)) \vert \; dt \ge 0. \tag 6$

We estimate the right-hand side of (4), using the definition

$\Vert f(x) \Vert_1 = \sup_{x \in [0, a]} \vert f(x) \vert: \tag 7$

$\displaystyle \int_0^a \vert t(y(t) - z(t)) \vert \; dt = \int_0^a \vert t \vert \vert y(t) - z(t) \vert \; dt$ $\le \displaystyle \int_0^a \vert t \vert \Vert y(x) - z(x) \Vert_1 \; dt = \Vert y(x) - z(x) \Vert_1 \int_0^a \vert t \vert \; dt; \tag 8$

note we can bring $\Vert y(x) - z(x) \Vert_1$ out of the integral since, being defined as a supremum over the set $[0, a]$, it is independent of $t$; finally,

$\displaystyle \int_0^a \vert t \vert \; dt = \int_0^a t \; dt = \dfrac{a^2}{2}; \tag 9$

therefore, (8) becomes

$\Vert (Ty)(x) - (Tz)(x) \Vert_1 = \displaystyle \int_0^a \vert t(y(t) - z(t)) \vert \; dt \le \dfrac{a^2}{2} \Vert y(x) - z(x) \Vert_1. \tag 9$

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    $\begingroup$ This is basically what I came up with, except I hesitated in (8) when taking the norm outside of the integral. Could you explain to me why this is possible? I might be misunderstanding this, but it seems to me that this is the equivalent of saying that $\int t(y(t)-z(t)) dt = \int t dt \int (y(t)-z(t)) dt$. Is that a property of integrals? It's certainly not one that I was aware of, if it is. $\endgroup$ – user569579 Oct 1 '18 at 20:04
  • $\begingroup$ @user569579: one can take the norm $\Vert y(t) - x(t) \Vert_1$ out of the integral because it really doesn't depend on $t$, hence is a constant with respect to the integration. Actually, my notation here was less than optimal; I should likely have used $\Vert y(x) - z(x) \Vert_1$ since then no $t$-dependence is suggested. I'll mos' likely edit my post . . . thanks. $\endgroup$ – Robert Lewis Oct 1 '18 at 20:13

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