functional analysis - Inequality involving supremum norm on integral

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.

2 Answers

  1. Leo- Reply

    2019-11-14

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

  2. Leonard- Reply

    2019-11-14

    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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