I believe the property that it doesn't follow is that ||f(x)|| = 0 doesn't always imply f(x) = 0
hmm
There are functions that are nonzero but have zero L2 norm - take any function that nonzero at one point and zero elsewhere. If we restrict ourselves to functions on the unit interval whose squares are continuous, this problem should go away.
Lavender
I agree with Fjorge and hmm about why L2 norm is a Pseudo-norm.
I believe the property that it doesn't follow is that ||f(x)|| = 0 doesn't always imply f(x) = 0
There are functions that are nonzero but have zero L2 norm - take any function that nonzero at one point and zero elsewhere. If we restrict ourselves to functions on the unit interval whose squares are continuous, this problem should go away.
I agree with Fjorge and hmm about why L2 norm is a Pseudo-norm.