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

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".

• It seems to me that you need to impose some further conditions on the constants, e.g. some control of the sharpness of the constants, because the constants that give you these inequalities are trivially not unique. For example, you can take the max of both $C$-s and the min of both $c$-s, and these will be "universal" with respect to both choices of function triplets.
– M.G.
Jul 1 at 18:29
• You are kinda still faced with the same issue, though. Regardless, if the same constants hold for any $n$-dimensional vector subspace, where $n$ is a fixed positive integer (in your case $n=3$), then they will trivially hold for any element of your infinite-dimensional vector space since in particular it's an element of (infinitely many) $n$-dimensional vector subspaces. Therefore the norms are going to be equivalent as norms on the infinite-dimensional vector space, a contradiction.
– M.G.
Jul 1 at 20:20
• $f(0)=f(1)$, this is what I meant. Jul 2 at 0:57
• Re, but these are $L^2$ functions, defined only "up to equality almost everywhere", so the condition $f(0) = f(1)$ from one point of view is always satisfied (for some representatives), and from another point of view is meaningless (because it is not independent of the choice of representative). Jul 2 at 2:17
• @PietroMajer, re, isn't that the same as @‍M.G.'s comment and answer? Jul 2 at 16:07

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.