# Strictly convex absolutly 1 homogeneous function

Is every strictly convex, 1-homogeneous function on $\mathbb R^d$ simply a multiple of the Euclidean norm?

Update: The above is no, since any p-norm on $\mathbb R^d$ is strictly convex and 1-homogeneous.

My new question is: Given a norm on $\mathbb R^d$ that is strictly convex, is there any way of characterizing it in relation to known norms? I.e. It's true if and only if the norm is a p-norm for $1< p <\infty$.

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It can't be strictly convex and positive homogenous. For example, if $f$ is positive homogeneous, then for $\lambda >0$, $f(\lambda x) = \lambda f(x)$, and $f(x)= \frac{1}{2}(f(\frac{1}{2}x)+f(\frac{3}{2}x))$, hence it cannot be strictly convex. Do you mean that the unit ball is strictly convex? – copper.hat Feb 1 '13 at 17:56
for your new question, even if the norm is not a convex function, one can bound one norm with another norm (upper and lower) for suitable constants. The proof escapes me now. – dineshdileep Feb 1 '13 at 19:53