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".



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


