# Showing linearity of functionals on $c[a,b]$ and $l^2$.

Erwin Kreyszig Section 2.8, Problem 1:

Define a functional on $C[a,b]$ by fixing $t_0\in[a,b]$ and setting:

$$f_1(x)=x(t_0)$$

Define a second functional on $l^2$ by choosing a fixed $a=(\alpha_j)\in l^2$ and setting $$f(x) = \sum_{j=1}^\infty \xi_j\alpha_j$$ where $x=(\xi_j)\in l^2$

Show these two functionals are linear

This question has been self-answered by the OP(me).

• If you have found this at some point. It may be of interest that Q2, Q3, Q4 from this section of the textbook have answers on this website. – Functional Analysis Oct 27 '15 at 17:34

For $1)$ We want to show this functional is linear.
I.e. $f(ax_1+bx_2)=af(x_1)+bf(x_2)$
$f(ax_1+bx_2) = (ax_1+bx_2)(t_0)=(ax_1)(t_0) + (bx_2)(t_0)=ax_1(t_0)+bx_2(t_0)=af(x_1)+bf(x_2)$$\square For 2) We want to show this functional is linear: I.e. \displaystyle\sum_{j=1}^\infty \left(m\xi_j\alpha_j + n\zeta_j\alpha_j \right)=f(mx_1+nx_2)=mf(a)+nf(b) = m\sum_{j=1}^\infty \xi_j\alpha_j+n\sum_{j=1}^\infty \zeta_j\alpha_j We have:$$\displaystyle\sum_{j=1}^\infty \left(m\xi_j\alpha_j + n\zeta_j\alpha_j \right)=\displaystyle\sum_{j=1}^\infty \left(m\xi_j\alpha_j\right) + \sum_{j=1}^\infty\left(n\zeta_j\alpha_j \right)=m\displaystyle\sum_{j=1}^\infty \left(\xi_j\alpha_j\right) + n\sum_{j=1}^\infty\left(\zeta_j\alpha_j \right)$\$