A question in Functional Analysis.

Question

I want to prove the following statement:

$\Omega$ is a compact set. Let $A_N $ be an $N$ dimensional subspace of $L_\infty(\Omega)$. $\{g_i\}_{i=1}^N$ be any orthonormal basis of $A_N\subset L_2(\Omega)$. Prove that for any $f=\sum_{i=1}^Nc_ig_i(x)$ , ($c_1,...,c_N\in \mathbb{R}$): $$\sup_{f\in A_N,f\neq0}\frac{|f(x)|}{||f||_2}=\sup_{(c_1,...,c_N)\in\mathbb{R}^N\\\sum_{i=1}^Nc_i^2=1}\big|\sum_{i=1}^Nc_ig_i(x)\big|.$$

My understanding is that we transfer the characterization of the norm on $f$ to the coefficients ${c}_i$ by considering a unit sphere: $\sum_{i=1}^N c_i^2=1$. But I have no idea to show this identity.

Any suggestions would be welcome! Thank you!

Answer 1

The equation simply says that $\sup \{\frac {\{|\sum c_ig_i(x)|} {|\sqrt {\sum |c_i|^{2}}}\}=\sup \{|\sum c_ig_i(x)|: \sum |c_i|^{2}=1\}$ where the supremum on the left is taken over all choices of $c_1,c_2....,c_N$. [I have used the orthonormlaity of $g_i$'s to get $\|f\|_1^{2}=\sum |c_i|^{2}$]. Now this identity has noting to do with $L^{2}$ functions. For any real numbers $a_1,a_2,..,a_N$ we have $\sup \{\frac {\{|\sum c_ia_i|} {|\sqrt {\sum |c_i|^{2}}}\}=\sup \{|\sum c_ia_i|: \sum |c_i|^{2}=1\}$ and this is elementary. Can you prove this? [I will post a detailed answer if you cannot see why this is true].

Answer 2

• Show that the left-hand side can be rewritten as $$\sup_{f \in A_N, \|f\|_2=1} |f(x)|.$$
• The supremum over functions $f \in A_N$ can be instead over functions of the form $\sum_{i=1}^N c_i g_i(x)$.
• Show that then the $\|f\|_2=1$ constraint then becomes $\|\sum_{i=1}^n c_i g_i\|_2^2 = \sum_{i=1}^N c_i^2 = 1$.