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Let $a$, $b$ be two real numbers such that $a<b$, and let $C[a,b]$ denote the normed space of all (real- or complex-valued) functions defined and continuous on the closed interval $[a,b]$ with the norm defined as $$\Vert x \Vert_{C[a,b]} \colon= \max_{t\in[a,b]} \vert x(t) \vert \text{ for all } x \in C[a,b].$$

Let $f_1$ and $f_2$ be two functionals defined on $C[a,b]$ as follows: $$f_1(x) \colon= \int_a^b x(t) y_0(t) dt \text{ for all } x \in C[a,b], \text{ where } y_0 \in C[a,b] \text{ is fixed.} $$ And $$f_2(x) \colon= \alpha x(a) + \beta x(b) \text{ for all } x \in C[a,b], \text{ where } \alpha, \beta \text{ are some fixed scalars.}$$

Then what is the value of $\Vert f_1 \Vert$ and $\Vert f_2 \Vert$?

I know that $f_1$ and $f_2$ are both linear.

And, for each $x \in C[a,b]$, we have $$\begin{aligned}\vert f_1(x) \vert & = \left\vert \int_a^b x(t) y_0(t) dt \right\vert \\ & \leq \int_a^b \vert x(t) \vert \cdot \vert y_0(t) \vert dt \\ & \leq \int_a^b \left( \max_{\tau\in[a,b]} \vert x(\tau) \vert \right) \cdot \vert y_0(t) \vert dt \\ & = \int_a^b \Vert x \Vert_{C[a,b]} \cdot \vert y_0(t) \vert dt \\ & = \Vert x \Vert_{C[a,b]} \int_a^b \vert y_0(t) \vert dt. \end{aligned}$$ Thus it follows that $f_1$ is bounded and that $$\Vert f_1 \Vert \leq \int_a^b \vert y_0(t) \vert dt.$$

And, for each $x \in C[a,b]$, we also have $$\begin{aligned} \vert f_2(x) \vert & = \vert \alpha x(a) + \beta x(b) \vert \\ & \leq \vert \alpha \vert \vert x(a) \vert + \vert \beta \vert \vert x(b) \vert \\ & \leq \left( \vert \alpha \vert + \vert \beta \vert \right) \max_{t\in[a,b]} \vert x(t) \vert \\ & = \left( \vert \alpha \vert + \vert \beta \vert \right) \Vert x \Vert_{C[a,b]}, \end{aligned}$$ which shows that $f_2$ is bounded and that $$\Vert f_2 \Vert \leq \left( \vert \alpha \vert + \vert \beta \vert \right).$$

What next? How to derive the reverse inequality in each case (if it holds in either case, that is!)?

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For the second case, write $\alpha=r e^{i \theta},\beta=\rho e^{i \phi}$, and let $x(t)$ be a linear function starting at $x(a)=e^{-i \theta}$ and terminating at $x(b)=e^{-i \phi}$.

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  • $\begingroup$ I didn't get your answer. So could you please write it in detail enough? $\endgroup$ – Saaqib Mahmood Mar 29 '15 at 14:53

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