Problem. Describe a norm $\lVert\cdot\rVert_0$ on $\mathbb{R}^3$ such that the standard unit vectors have norm $1$ while $\lVert(1,1,1)\rVert_0<\frac{1}{100}$.
This norm looks not so trivial because of the restriction $\lVert(1,1,1)\rVert_0<\frac{1}{100}$.
$p$-norms do not work at all. Even though $\lVert(1,1,1)\rVert_p$ decreases as $p\to\infty$, we still only have $\lVert(1,1,1)\rVert_\infty=1\gg 1/100$.
Hope anyone have good ideas.
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$\begingroup$The first approach I thought of was as follows: for any positive definite matrix $A$, the function $\|x\|_A = \sqrt{x^TAx}$ defines a norm, and the diagonal entries of $A$ are the norms of the standard unit vectors.
With that in mind, consider the matrix$$ A = \pmatrix{1 & \epsilon - 1/2& \epsilon - 1/2\\\epsilon - 1/2 & 1 & \epsilon - 1/2 \\ \epsilon - 1/2 & \epsilon - 1/2 & 1}, $$where $0 < \epsilon < 1/2$. The corresponding norm satisfies$$ \|x\|_A^2 = \sum_{i=1}^3 x_i^2 - \sum_{1 \leq i<j \leq 3} (1 - 2 \epsilon) x_ix_j. $$We find that the standard unit vectors have norm $1$ for all such $\epsilon$. However, for sufficiently small $\epsilon$, we find that $\|(1,1,1)\|_A = \sqrt{6\cdot \epsilon} < 1/100.$
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