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A vector norm (in $\mathbb{R}^n$) is just a function $f:\mathbb{R}^n\to\mathbb{R}$ satisfying certain properties. If you put the positive homogeneity property together with the triangle inequality you get convexity of $f$: Let $\alpha\in[0,1]$, and $x,y\in\mathbb{R}^n$. Then $$f(\alpha x+(1-\alpha)y)\leq f(\alpha x)+f((1-\alpha)y)=\alpha f(x)+(1-\alpha)f(y)... 1 Hint: For any x with \|x\|_1=1, we have \|x\|_\infty \leq 1. 1 Lets try the layman's approach:$$\vec v=\begin{bmatrix}a+b\\a-b\\a+b\\a-b\end{bmatrix}\ \ (a,b\in \Bbb{R})$$We need to find$$\min_{a,b}||\vec v-\vec w||$$Check that the values of a,b for which ||\vec v-\vec w|| will be minimum = values of a,b for which ||\vec v-\vec w||^2 will be minimum. Now$$\min_{a,b}||\vec v-\vec w||^2\\=\min_{a,b}\{(a+b-1)...

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Here's a proof why $l^p(\mathbb N)$ is not locally convex, this is just for simplicity, it can be easily generalized. If it would be locally convex, then the unit ball $B_1(0)$ would contain a convex neighborhood U of $0$. Then there must be $\delta>0$ with $B_{2\delta}(0)\subset U$, hence also $\mathrm{conv}(B_{2\delta}(0))\subset U\subset B_1(0)$. Let ...

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Your transition from the first line to the second is incorrect. We should have  \|x - \alpha\|^2 - \|x - \beta\|^2 = \\ (x - \alpha)^T(x - \alpha) - (x - \beta)^T(x - \beta) = \\ \|x\|^2 + \|\alpha\|^2 - \|x\|^2 - \|\beta\|^2 - x^T\alpha - \alpha^Tx + x^T\beta + \beta^Tx =\\ \|\alpha\|^2 - \|\beta\|^2 - 2\alpha^Tx + 2 \beta^Tx =\\ \alpha^T\alpha - \beta^T\...

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