Linear Functionals: Why Not ##I(u) = \int_a^b u\frac{du}{dx}dx##?

Ok, so now I am confused. The book clearly states that a functional is said to be linear if ##l(\alpha u+\beta v)=\alpha l(u)+\beta l(v)##...but this does not seem to be the case for the second example I gave. So is the book incorrect? Or am I missing something?In summary, a functional is said to be linear if it satisfies the relation ##l(\alpha u+\beta v)=\alpha l(u)+\beta l(v)##. However, in the example given, ##l(u)=\displaystyle \int_a^b f(x) u dx +c##, this relation does not hold true. Therefore, the book may be incorrect
  • #1
bugatti79
794
1

Homework Statement



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##

Homework Equations


where ##\alpha## and ##\beta## are real numbers and ##u## , ##v## are dependant variables.

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
 
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  • #2
[tex] I(\alpha u) = \int_a^b \alpha u \frac{d (\alpha u)}{d x} dx [/tex]
 
  • #3
Very cursorily, it is not "linear" because it is a product of two terms involving u.

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

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

[tex]= \int_a^b (\alpha u+ \beta v)(\alpha\frac{du}{dx}+ \beta\frac{dv}{dx})dx[/tex]
[tex]= \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[/tex]
 
  • #4
The_Duck said:
[tex] I(\alpha u) = \int_a^b \alpha u \frac{d (\alpha u)}{d x} dx [/tex]

HallsofIvy said:
Very cursorily, it is not "linear" because it is a product of two terms involving u.

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

HallsofIvy said:
Your formula for [itex]l(\alpha u+ \beta v)[/itex] is incorrect. You are doing as if [itex](a+ b)^2[/itex] were equal to [itex]a^2+ b^2[/itex] and that is not true.

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

[tex]= \int_a^b (\alpha u+ \beta v)(\alpha\frac{du}{dx}+ \beta\frac{dv}{dx})dx[/tex]
[tex]= \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[/tex]

...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
Well, as I showed in my first response, it is NOT [itex]\alpha l(u)+ \beta l(v)[/itex],
it is [itex]\alpha^2l(u)+ \beta^2l(v)[/itex] plus two additional terms!
 
  • #6
bugatti79 said:
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?

HallsofIvy said:
Well, as I showed in my first response, it is NOT [itex]\alpha l(u)+ \beta l(v)[/itex],
it is [itex]\alpha^2l(u)+ \beta^2l(v)[/itex] plus two additional terms!

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
Nonononononono!

First of all [itex]l(v)=\displaystyle\int\limits_a^b f(x)vdx+c[/itex], NOT [itex]l(v)=\displaystyle\int\limits_a^b g(x)vdx+d[/itex]. Who give you rights to say [itex]g(x)[/itex] or [itex]d[/itex]? NOBODY!
[itex]l(\alpha u + \beta v) = \displaystyle\int\limits_a^b f(x)(\alpha u + \beta v)dx + c[/itex], not [itex]+\alpha c + \beta c[/itex]
 
  • #8
Karamata said:
Nonononononono!

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

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, that's not in the relation?
 
  • #9
bugatti79 said:
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, that's not in the relation?

This isn't good, because

[itex]\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[/itex]
 
  • #10
Karamata said:
This isn't good, because

[itex]\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[/itex]

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

1. What is a linear functional?

A linear functional is a mathematical function that takes in a vector as input and outputs a scalar value. It can be thought of as a generalization of the dot product between two vectors.

2. How is a linear functional different from a regular function?

A regular function takes in a scalar value as input and outputs a scalar value, while a linear functional takes in a vector as input and outputs a scalar value. Additionally, a linear functional must satisfy the properties of linearity, meaning that it follows the rules of addition and scalar multiplication.

3. What is the significance of the notation ##I(u) = \int_a^b u\frac{du}{dx}dx## in linear functionals?

This notation represents the linear functional that takes in a function u as input and outputs the integral of u multiplied by its derivative over the interval [a,b]. This notation is often used in the context of functional analysis, a branch of mathematics that studies spaces of functions and their properties.

4. Why is the notation ##I(u) = \int_a^b u\frac{du}{dx}dx## used instead of a simpler form?

The use of this notation is a convention in functional analysis, as it allows for the representation of more complex linear functionals in a concise and standardized form. It also highlights the connection between the integral and derivative in the definition of a linear functional.

5. How are linear functionals used in practical applications?

Linear functionals have various applications in mathematics, physics, and engineering. They are used in functional analysis to study the properties of function spaces, in optimization problems to find optimal solutions, and in physics to describe physical systems using mathematical models.

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