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?

  • $\begingroup$ @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

Your attempt is fine, everything is O.K., no errors !


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.