Determining whether something is a valid norm in linear algebra

trobinson41

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Dec 29, 2014
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Let C[a,b] be the set of all real-values functions that are continuous on the closed interval [a,b]. According to my book the following does NOT define a norm:
|| f || = |f(a)| + |f(b)|

The book does not explain why. I'm guessing that it doesn't satisfy the triangle inequality, but I can't think of a counterexample.

A norm has to satisfy the following conditions:
Let v, w be members of a vector space V.
1. || v || >= 0 and is equal to 0 if and only if v = 0.
2. || av || = |a| || v || for any scalar a.
3. || v + w || <= || v || + || w ||

Thanks.
 
Let C[a,b] be the set of all real-values functions that are continuous on the closed interval [a,b]. According to my book the following does NOT define a norm:
|| f || = |f(a)| + |f(b)|

The book does not explain why. I'm guessing that it doesn't satisfy the triangle inequality, but I can't think of a counterexample.

A norm has to satisfy the following conditions:
Let v, w be members of a vector space V.
1. || v || >= 0 and is equal to 0 if and only if v = 0.
2. || av || = |a| || v || for any scalar a.
3. || v + w || <= || v || + || w ||

Thanks.

It satisfies the triangle inequality
||v + w|| = |v(a) + w(a)| + |v(b) + w(b)| \(\displaystyle \le\) |v(a)| + |w(a)| + |v(b)| + |w(b)| = ||v|| + ||w||
but isn't
f(x) = (x-a) (b-x)
continuous on [a,b] and non-zero on (a,b).
 
It satisfies the triangle inequality||v + w|| = |v(a) + w(a)| + |v(b) + w(b)| \(\displaystyle \le\) |v(a)| + |w(a)| + |v(b)| + |w(b)| = ||v|| + ||w||but isn'tf(x) = (x-a) (b-x)continuous on [a,b] and non-zero on (a,b).
I think that's the answer. Thanks for your help
:razz:
 
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