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Let $N$ be a norm on $\mathbb R^2$, such that $N ( \mathbb Z^2) \subset \mathbb N $, where $\mathbb Z^2 =\{ (a,b)\mid a\mbox{ and }b \mbox{ are integers}\}$.

Help me to prove that for $u$, $v$ fixed vectors the following limit exists

$$\lim_{ t\to 0^+} \frac{N(u+tv)-N(u)}t. $$

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Did you try to use the convexity of the norm? –  Davide Giraudo Dec 18 '11 at 10:31
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1 Answer 1

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I don't see where your additional assumption should be used. Are you trying to prove some stronger claim?

As pointed out in Davide Giraudo's comment, the key property to use here is convexity of the norm.

Let $X$ be any linear normed space.

For any $a\in[0,1]$ and $x,y\in X$ we get $$N(ax+(1-a)y) \le N(ax)+N((1-a)y)=aN(x)+(1-a)N(y),$$ i.e. norm is a convex function.

If we denote $g(t)=N(u+tv)$, then for any $a\in[0,1]$ and $x,y\in\mathbb R$ \begin{align*}g(ax+(1-a)y)&=N(u+axv+(1-a)ayv)\\ &=N(au+axv+(1-a)u+(1-a)ayv)\\ &=N(a(u+xv)+(1-a)(u+yv)) \\ &\le aN((u+xv))+(1-a)N(u+yv)\\ &=ag(x)+(1-a)g(y), \end{align*} i.e. $g$ is a convex real function.

The limit in the questions is precisely the right derivative of $g$ at the point $0$. It is known that every convex function has one-sided derivatives at each point, see e.g. Radulescu, Radulescu, Andreescu: Problems in real analysis, Problem 6.3.5, p.276 or Dudley: Real analysis and probability, Corollary 6.3.3., p.204. (And probably almost any textbook in real analysis, I've included simply the books which were among the first results when searching for convex derivative "one-sided" "real analysis").

Differentiability of norm (in the sense of directional, Frechet or Gateaux derivative) seems to be studied in many books. E.g. the following might be interesting for you: Pavel Drábek, Jaroslav Milota: Methods of nonlinear analysis, Example 3.2.9, p.120. The first part of that example answers your question, the argument is the same as I've given above; I've tried to include more details.

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