- #1
bugatti79
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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