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In our class of functional analysis, we only studied theory, but never did exercises. We haven't studied Lebesgue integration. Also I'm having problems with key understanding of the subject, please don't judge me for that.

Let $A : C[a,b] \mapsto \mathbb{R}$ be a functional, defined as $A(x) = \int_{a}^{b}x(t) t^2 dt$.

1) Show that it's linear and bounded.
2) Compute it's norm.

To show that it's bounded I tried this approach. $$||A(x)|| = \sup_{t\in [a,b]} \left\lvert \int_{a}^{b}x(t) t^2 dt \right\rvert \leq \sup_{t\in [a,b]} \int_{a}^{b} |t^2| \sup_{t\in [a,b]} |x(t)| dt = \sup_{t\in [a,b]} \int_{a}^{b} |t^2| ||x|| dt = ||x|| \sup_{t\in [a,b]} {\int_{a}^{b} |t^2| dt} = sgn(|b|-|a|)\frac{(b^2 - a^2)}{2}||x|| $$ So $||A(x)|| \leq C ||x||$

For linearity I think I need to show $A(\alpha x + \beta y) = \alpha A(x) + \beta A(y), \alpha, \beta \in \mathbb{R}, x,y \in C[a,b]$.
This follows from linearity property of integral.

To compute the norm of this operator, I did some searching at this site and saw that most of approaches here look like this.

1) showing that $||A(x)|| \leq C||x||$
2) showing that $||A(x)|| \geq C||x||$
conclude that $||A|| = C$

Will this approach be correct to continue? I will need some advice if my steps were wrong, and/or some hints to continue. Thank you.

shcolf
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  • Right after the equals sign to the right of $| A (x) |$ (the first time it appears), you cannot take the supremum over $t$. $t$ is just a dummy variable here. The definition of the norm is a supremum over $x$. – Theoretical Economist Dec 21 '16 at 10:29

1 Answers1

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We can assume that $a<b$. Then

$|A(x)| \le \int _{a}^{b}|x(t)|t^2 dt \le ||x||\int _{a}^{b}t^2 dt=\frac{b^3-a^3}{3}||x||$.

Hence $||A || \le \frac{b^3-a^3}{3}$.

With $x(t)=1$ for all $t \in [a,b]$, we have $|A(x)|=\frac{b^3-a^3}{3}$. hence

$$||A || = \frac{b^3-a^3}{3}.$$

Fred
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  • Thanks a lot sir! I've done stupid mistake while computing that integral. I will accept your answer. If you spotted any conceptual mistake in my thoughts, would be great if you tell me about that. – shcolf Dec 21 '16 at 10:10