Norm equivalence in finite dimensions - is the equivalence "universal" if the dimension is fixed?

Question

I am aware that in a finite dimensional vector space, any two norms are equivalent.

However, I cannot really figure out how "universal" the equivalence constants are.

To be specific, let us think of the space $L^2\Bigl([0,1],\mathbb{R} \Bigr)$ of periodic real-valued functions on $[0,1]$. Denote its inner product by $\langle , \rangle_{L_2}$.

Let us choose "any triplet" of "smooth" functions $f_1, f_2, f_3 : [0,1] \to \mathbb{R}$ which are orthonormal with resept to $\langle , \rangle_{L_2}$ and denote their linear span as \begin{equation} V_{f_1,f_2,f_3}:= \langle f_1, f_2, f_3 \rangle \subset L^2\Bigl([0,1],\mathbb{R} \Bigr). \end{equation} Then, $V_{f_1,f_2,f_3}$ is clearly just $3$-dimensional and therefore all norms must be equivalent on it.

That is, even the Sobolev $H^1$ norm, denoted as $\lVert \cdot \rVert_{H_1}$ must be equivalent to the $L^2$ norm $\lVert \cdot \rVert_{L^2}$, which means that there are constants $c, C>0$ satisfying \begin{equation} c \lVert f \rVert_{H_1}\leq \lVert f \rVert_{L^2} \leq C \lVert f \rVert_{H_1} \end{equation} for all $f \in V_{f_1,f_2,f_3}$.

Now, my question is that, are these constants $c$ and $C$ "universal"?

That is, if I choose another triplet of $L^2$-orthonormal smooth functions $g_1, g_2, g_3$ entirely different from $f_1, f_2, f_3$ above and consider $V_{g_1,g_2,g_3}$, do we still have \begin{equation} c \lVert g \rVert_{H_1}\leq \lVert g \rVert_{L^2} \leq C \lVert g \rVert_{H_1} \end{equation} for all $g \in V_{g_1,g_2,g_3}$ and the same constants $c, C$?

OK I will clarify my question: I wonder if there exists a pair $(c,C)$ validating the above equivalence relation for "any" choice of $L^2$-orthonormal triplet of smooth functions. In this sense, the pair $(c,C)$ is "universal".

Answer 1

I am turning my comment into an answer.

If I've understood your question correctly, the constants $(c,C)$ cannot be "universal" in your sense for rather trivial set-theoretical reasons.

Suppose that the norm inequality is valid for every 3-dimensional vector subspace for the same constant choice of constants $(c,C)$, where 3 really is an arbitrary fixed positive integer. Then, since every element of the infinite-dimensional vector space is in particular an element of a 3-dimensional vector subspace, the norm inequality with $(c,C)$ is valid on the whole infinite-dimensional vector space. Hence both norms must be equivalent as norms on the whole infinite-dimensional vector space.

In other words, two inequivalent norms can't become equivalent with the same constants on all $n$-dimensional vector subspaces.