# Show that the norm of the derivitive of a $C^1$ function over a vector space is non-negative, homogeneous and satisfies the triangle ineq

For $f$ in $C^1[a,b]$, define $p(f)= \parallel f'\parallel _{\infty}$. Show that $p$ is non-negative, homogeneous, and satisfies the triangle inequality. Why is it not a norm?

-I can easily show the supremum of the norm of a function continuous once is positive and homogeneous but im not sure how to show the triangle inequality holds and have no clue why it is not a norm.

Any guidance would be appreciated

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It's not a norm since $p(C) = 0$ for any constant function. For the triangular inequality, $$p(f+g) = \|f'+g'\|_\infty = \sup_x |f'(x)+g'(x)|\leq \sup_x |f'(x)| + \sup_y|g'(y)| = p(f) + p(g)$$ – Ilya Jan 31 '13 at 18:25
@Ilya oh i see, darn that didn't even cross my mind – bobdylan Jan 31 '13 at 18:40