Proving Another Vector Norm on C[0,1]

[gtfitzpatrick]

Homework Statement

does the function $$\| \|: C[0,1] \rightarrow$$ R defined by $$\|f \|= |f(1)- f(0)|$$ define a norm on C[0,1]. if it does prove all axioms if not show axiom which fails

The Attempt at a Solution

i don't really understand the question. i know the 4 axioms of a norm but i don't know how to use the info given to prove or disprove them. is the question telling me that x has to be either 0 or 1 and y has to be either 0 or 1.

 

[gtfitzpatrick]

wait so its actually telling me its a straight line at y=1 between the points x=0 and x=1. so f(1) = 1 and f(0) = 1

so axiom 2 fails $$<br /> \|f \| = 0<br />$$ iff f = $$\vec 0$$ but $$<br /> \|f \|= |f(1)- f(0)| = |1- 1| = 0 <br />$$

 

[ystael]

Your words and reasoning seem garbled, but your counterexample is correct: $$\|f\| = |f(1) - f(0)|$$ does not define a norm on $$C[0,1]$$, because if $$f(x) = 1$$ is the constant function then $$\|f\| = |1 - 1| = 0$$ even though $$f \neq 0$$.