Linear Functionals

1. Jul 11, 2012

bugatti79

1. The problem statement, all variables and given/known data

Why does this not qualify as a linear functional based on the relation $l(\alpha u+\beta v)=\alpha l(u)+\beta l(v)$?

$\displaystyle I(u)=\int_a^b u \frac{du}{dx} dx$

2. Relevant equations
where $\alpha$ and $\beta$ are real numbers and $u$ , $v$ are dependant variables.
3. The attempt at a solution
If we let $\displaystyle I(v)=\int_a^b v \frac{dv}{dx} dx$

then $l(\alpha u+\beta v)=$

$\displaystyle \int_a^b ( \alpha u \frac{du}{dx} dx + \beta v \frac{dv}{dx} dx)=\displaystyle \int_a^b \alpha u \frac{du}{dx} dx +\int_a^b \beta v \frac{dv}{dx} dx=\displaystyle \alpha\int_a^b u \frac{du}{dx} dx +\beta \int_a^b v \frac{dv}{dx} dx=\alpha l(u)+\beta l(v)$...? THanks

2. Jul 11, 2012

The_Duck

$$I(\alpha u) = \int_a^b \alpha u \frac{d (\alpha u)}{d x} dx$$

3. Jul 11, 2012

HallsofIvy

Staff Emeritus
Very cursorily, it is not "linear" because it is a product of two terms involving u.

Your formula for $l(\alpha u+ \beta v)$ is incorrect. You are doing as if $(a+ b)^2$ were equal to $a^2+ b^2$ and that is not true.

You need
$$l(\alpha u+ \beta v)= \int_a^b (\alpha u+ \beta v)\frac{d(\alpha u+ \beta v)}{dx}dx$$

$$= \int_a^b (\alpha u+ \beta v)(\alpha\frac{du}{dx}+ \beta\frac{dv}{dx})dx$$
$$= \alpha^2 \int_a^b u\frac{du}{dx}dx+ \alpha\beta\int_a^b u\frac{dv}{dx}dx+ \alpha\beta\int_a^b v\frac{du}{dx}dx+ \beta^2\int_a^bv\frac{dv}{dx}dx$$

4. Jul 11, 2012

bugatti79

This is what I thought so too and that this non linearity has nothing to do with the realtion in my first thread...however

......in the book (which I am self studying finite element theory) it states

" a functional $l(u)$ is said to be linear in u iff it satisfies the relation..."

$l(\alpha u+\beta v)= \alpha l(u)+\beta l(v)$.....? How is this wrong?

5. Jul 11, 2012

HallsofIvy

Staff Emeritus
Well, as I showed in my first response, it is NOT $\alpha l(u)+ \beta l(v)$,
it is $\alpha^2l(u)+ \beta^2l(v)$ plus two additional terms!

6. Jul 12, 2012

bugatti79

Ok, what about this one. Using the same relation $l(αu+βv)=αl(u)+βl(v)$ for a functional $l(u)=\displaystyle \int_a^b f(x) u dx +c$? The book states this is not a linear functional...? Why? Heres my attempt..

let $l(v)=\displaystyle \int_a^b g(x) v dx +d$ then the LHS of the relation can be written as

$\displaystyle \int_a^b \alpha f(x) u dx +\alpha c + \displaystyle \int_a^b \beta g(x) v dx +\beta d=\alpha (\int_a^b f(x) u dx +c) + \beta (\displaystyle \int_a^b g(x) v dx +d)=αl(u)+βl(v)$...Why is this not a linear functional?

7. Jul 12, 2012

Karamata

Nonononononono!

First of all $l(v)=\displaystyle\int\limits_a^b f(x)vdx+c$, NOT $l(v)=\displaystyle\int\limits_a^b g(x)vdx+d$. Who give you rights to say $g(x)$ or $d$? NOBODY!
$l(\alpha u + \beta v) = \displaystyle\int\limits_a^b f(x)(\alpha u + \beta v)dx + c$, not $+\alpha c + \beta c$

8. Jul 12, 2012

bugatti79

ok...the the RHS would end up like
$\displaystyle \alpha \int\limits_a^b f(x)u dx + \beta \int\limits_a^b f(x)vdx + c=\alpha l(u)+\beta l(v)+c$

What about the c though, thats not in the relation?

9. Jul 12, 2012

Karamata

This isn't good, because

$\alpha l(u) + \beta l(v)+c = \alpha \left(\displaystyle\int\limits_a^b f(x)udx+c\right)+\beta \left(\displaystyle\int\limits_a^b f(x)vdx+c\right)+c = \alpha\displaystyle\int\limits_a^b f(x)udx + \alpha c + \beta \displaystyle\int\limits_a^b f(x)vdx + \beta c + c \neq \displaystyle \alpha \int\limits_a^b f(x)u dx + \beta \int\limits_a^b f(x)vdx + c$

10. Jul 12, 2012

bugatti79

So it cannot be a linear functional then...right? Looks good.THanks!!