Suppose that $\|\cdot\|$ is a norm on $\mathbb{R}^n$ and $\|\cdot\|^*$ is its dual norm. Consider two vectors $u$ and $v$ in $\mathbb{R}^n$ such that $u+v\neq 0$. Let $x$ be a vector that maximizes $x^T(u+v)$ subject to $\|x\|^* \le 1$, the optimal value being $\|u+v\|$ as we know.
If $\|\cdot\|$ is the $2$-norm, then $x= t(u+v)$ with a scalar $t>0$. Hence $$ x^T u -x^T v= t(u+v)^T(u-v) = t(\|u\|^2 -\|v\|^2), $$ which tells us that $$ x^T u \ge x^T v \quad \Longleftrightarrow \quad \|u\|\ge \|v\|. $$
Roughly speaking, if $u+v$ is closer to $u$ than to $v$, then $x^Tu$ is larger than $x^T v$ with this special $x$, and vice versa. Why is this interesting? Imagine that $u$ is a signal and $v$ is a perturbation. In some scenarios, we can observe $u+v$ only by the linear functional $x$. Naturally, we hope that the observation contains more truth than error provided that the signal is stronger than the perturbation. It turns out true if the geometry of $\mathbb{R}^n$ is the Eucleadian one. What if we consider other geometry?
If $\|\cdot\|$ is induced by an inner product, we still have $x^T u \ge x^T v \Longleftrightarrow \|u\|\ge \|v\|$, which can be proved by a similar argument.
However, the same implication does not hold if $\|\cdot\|$ is the $1$-norm. Consider the following vectors in $\mathbb{R}^2$: $$ u=(1,1)^T \quad \text{ and } \quad v = (-1-\epsilon, -1+2\epsilon)^T, $$ where $\epsilon\in(0,1/2)$. Then $u+v=(-\epsilon, 2\epsilon)^T$, and hence $x = (-1, 1)^T$. It is easy to check that $\|u\|=2>2-\epsilon = \|v\|$ whereas $x^T u = 0 < 3\epsilon = x^T v$.
Question. Under the setting specified in the first paragraph, can we guarantee the equivalence between $x^Tu\ge x^T v$ and $\|u\| \ge \|v\|$ for a class of norms more general than those induced by inner products?
An inference that may be wrong: by continuity, the abovementioned example for the $1$-norm may also work for $p$-norm with $p>1$ but close to $1$. Assuming this is true, the desired implication is not ensured if $\|\cdot\|$ is uniformly convex. Then maybe inner-product-induced norms are the only possible ones to ensure it. It would not be extraordinary if this turns out true because dot product and inner products are nothing but representations of linear functionals on $\mathbb{R}^n$ and they are naturally connected. A rigorous proof would be very intriguing. This is inspired by interactions with @supinf in the comments.
Even though it is not the primary objective, an extension to more general spaces will also be interesting --- doing so for inner-product spaces should not be difficult.
Any comments or criticism will be appreciated. Thanks.