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Let $a$ and $b$ be two real numbers such that $a < b$; let $\mathrm{C}[a, b]$ denote the normed space of all the (real or) complex-valued functions defined and continuous on the closed interval $[a, b]$ on the real line, with the maximum norm.

For each $x \in \mathrm{C}[a, b]$, let us define the function $\tilde{x}$ on $[a, b]$ as follows: $$ \tilde{x}(t) \colon= \int_a^t x(\tau) \ \mathrm{d} \tau \mbox{ for each } \tau \in [a, b]. $$ By Theorem 6.20, each $\tilde{x}$ is continuous (in fact, even differentiable) on $[a, b]$.

Am I right?

This defines a mapping $f \colon \mathrm{C}[a, b] \to \mathrm{C}[a, b]$, $x \mapsto \tilde{x}$, which is linear and bounded.

What is the norm of $f$?

My Attempt:

For every $x \in \mathrm{C}[a, b]$, we have $$ \lvert \tilde{x} (t) \rvert = \left\lvert \int_a^t x(\tau) \ \mathrm{d} \tau \right\rvert \leq \int_a^t \left\lvert x(\tau) \right\rvert \ \mathrm{d} \tau \leq \int_a^t \lVert x \rVert \ \mathrm{d} \tau = (t-a) \lVert x \rVert \leq (b-a) \lVert x \rVert. $$ and so $$ \lVert \tilde{x} \rVert = \max \{ \ \lvert \tilde{x}(t) \rvert \ \colon \ t \in [a, b] \ \} \leq (b-a) \lVert x \rVert, $$ which shoows that $f$ is indeed bounded, and upon taking the supremum over all $x$ of unit norm, we obtain $$ \lVert f \rVert \leq b-a. $$

Now for $x$ defined as $x(t) \colon= 1$ for all $t \in [a, b]$, we find that $\lVert x \rVert = 1$ and also that $$ \tilde{x}(t) = \int_a^t x(\tau) \ \mathrm{d} \tau = t-a, $$ and so $\lVert \tilde{x} \rVert = b-a$, from which it follows that $$ \lVert f \rVert \geq b-a. $$ Hence $\lVert f \rVert = b-a$.

Is what I have done so far correct? If not, then where lies the error?

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  • $\begingroup$ @user284331 what have you edited in my post? $\endgroup$ – Saaqib Mahmood Dec 15 '17 at 13:50
  • $\begingroup$ No, you just had a typo, $\displaystyle\int |x(\tau) d\tau$ and I removed the $|$. $\endgroup$ – user284331 Dec 15 '17 at 18:06
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Your attempt is fine, everything is O.K., no errors !

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