Question

Suppose that for each in P the function y is defined by
\[y (x)= \int_0^2 (x (t))^2dt\]
is y a linear functional?
Please, help

Answer 1

By definition, y is called a linear functional if it satisfy the condition
\[y(\alpha x_1+\beta x_2)= \alpha y(x_1)+\beta y(x_2)\]

Answer 2

My work: \[y (\alpha x_1) = \int_0^2 \alpha^2(x_1(t))^2 dt~~ (1)\]
I want to confirm this equation first. Am I right or
\[y (\alpha x_1) = \int_0^2 \alpha(x_1(t))^2 dt ~~(2)\]
which one is the right one (1) or (2)
then, I am OK

Answer 3

\[y (\alpha x_1) = \int_0^2 (\alpha x_1(t))^2 dt=\int_0^2 \alpha^2(x_1(t))^2 dt\]

Answer 4

what is P?

Answer 5

Thank you @Zarkon. I can step up from here. I don't want to go so far if I have a mistake there, :)
P is a polynomial

Answer 6

*set

Answer 7

ok

Answer 8

looks good

Answer 9

How about \[y(x) = \int_0^1 t^2x(t) dt\]

Answer 10

\[y(\alpha x_1(t) = \alpha^2t^2x_1(t) dt\]
right?

Answer 11

oh, I forgot \(\int_0^1\) in the front

Answer 12

yes. You are checking if this particular parametrized equations are linear, so yes. That's how you would check if functions scales.

Answer 13

not y(x(at)) but y(ax(t), like you did, yes.

Answer 14

what is difference between the 2?

Answer 15

oh, I got it, friend. like f (3x) \(\neq\) 3 f(x). right?

Answer 16

For example, if you ook at a line parametrized by arc length.

If S(t) is arclength (parametrized) then aS(t) is \(a\) times bigger than S(t). But S(at) is not \(a\) times bigger than S(t)

Answer 17

yep, that's it

Answer 18

got you :)

Answer 19

kool!

Answer 20

hey, I still have one more, hehehe. need check
y(x) = \(\dfrac{d^2x}{dt^2}|t=1\)
it looks not nice at all.

Answer 21

The derivative is a linear operator