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Suppose that we have a sequence of norms $(\|\cdot\|_n)_{n\geq 1}$ on a finite dimensional space $X$ such that, as functions from $X$ to $\mathbb{R}_+$, they point-wise converge to a limit norm $\|\cdot\|_\infty$. Since we are in a finite-dimensional setting, all of the norms are equivalent among them, and, in particular, all of the norms in the sequence are equivalent to $\|\cdot\|_\infty$. An implication of this fact is that there is a sequence of constants $c_n$ such that $$ \forall n \geq 1, \forall x\in X, \quad \|x\|_n \geq c_n\|x\|_\infty. $$

Now, the equivalence constants $c_n$ are not necessarily unique, but, since the norms $\|\cdot\|_n$ approach $\|\cdot\|_\infty$ in the limit, I am wondering how can one prove that they can be chosen to have $1$ as a limit (i.e. $\lim_{n\to\infty} c_n = 1$) ?

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Denote $$i_n = \inf\limits_{\Vert x \Vert_\infty = 1} \frac{\Vert x \Vert_n}{\Vert x \Vert_\infty}, s_n = \sup\limits_{\Vert x \Vert_\infty = 1} \frac{\Vert x \Vert_n}{\Vert x \Vert_\infty}$$ Those numbers are finite as $S=\{x \in X \ ; \ \Vert x \Vert_\infty = 1\}$ is compact and the norms are continuous on finite dimensional spaces.

Now, the norms $\Vert \cdot \Vert_n$ converge uniformly on $S$ as $S$ is compact. Therefore $(i_n)$ and $(s_n)$ converge to $1$, providing the result you were looking for.

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  • $\begingroup$ The uniform convergence on $S$ is the meat of the proof. For general continuous functions, pointwise convergence does not imply uniform convergence on compact subsets. $\endgroup$ – Daniel Fischer Dec 2 '16 at 14:26
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Essentially the question is whether pointwise convergence of a sequence of norms implies uniform convergence on a sphere. While for general continuous functions on a compact space, pointwise convergence to a continuous function does not imply uniform convergence, norms have more structure, and so in this setting we have uniform convergence on compact subsets of the space.

The key ingredient is that on a finite-dimensional real or complex vector space, the pointwise convergence of a sequence of norms implies equicontinuity of the sequence, and equicontinuity together with pointwise convergence implies locally uniform convergence.

Thus let's set out to prove equicontinuity. Choose a basis $x_1,\dotsc, x_d$ of $X$, and look at the norm

$$\Biggl\lVert \sum_{k = 1}^d a_k x_k\Biggr\rVert = \sum_{k = 1}^d \lvert a_k\rvert,$$

and its unit sphere $S = \{ x \in X : \lVert x\rVert = 1\}$. By the pointwise convergence, for $1 \leqslant k \leqslant d$ there is a constant $M_k \in (0,+\infty)$ with

$$\lVert x_k\rVert_n \leqslant M_k$$

for all $n$. With $M = \max \{ M_k : 1 \leqslant k \leqslant d\}$, the triangle inequality yields

$$\Biggl\lVert\sum_{k = 1}^d a_kx_k\Biggr\rVert_n \leqslant \sum_{k = 1}^d \lvert a_k\rvert \lVert x_k\rVert_n \leqslant M\sum_{k = 1}^d \lvert a_k\rvert,$$

and thus $\lVert x\rVert_n \leqslant M\lVert x\rVert$ for all $x$. Thus the family $(\lVert\cdot\rVert_n)_{n\in\mathbb{N}}$ is (uniformly) equicontinuous.

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