Let $V$ be a finite dimensional subspace of $C[0,1]$. Prove that there exists $g \in V$ such that $\|g\|_2 =1$, $\|g\|_\infty^2 \ge \dim V$
In this problem we use the notation $$\|f\|_2 = \left(\int_0^1 f^2\right)^{1/2}$$ $$\|f\|_\infty = \max_{[0,1]} f$$
My attempt
For the n-dimensional situation. Suppose that $f_i=\chi_{[\frac{i-1}{n}, \frac{i}{n}]}$ are the characteristic functions of intervals.
Then the condition is equivalent to that for $f=\sum_{i=1}^n \lambda_i f_i$ $$\sum_{i=1}^n \lambda_i^2 = n^2$$
Thus $$\|f\|_\infty^2 = \max \lambda_i^2 \geqslant n$$The inequality holds.
But I got stuck on analyzing the general situation. Could you please give me some hints? Thanks in advance!
Here is an answer. The core idea is to find an orthogonal basis to reduce complexity.
Proof
Choose the orthogonal basis $f_1,f_2,\cdots,f_n$ such that $$ \int_0^1f_if_j=\delta_{ij} \quad \forall\, 1\le i \le j\le n $$ Thus for $f=\sum_{i=1}^n \lambda_i f_i$ $$ \int_0^1 f^2 = \int_0^1 \sum_{i=1}^n \lambda_i^2 f_i^2 = \sum_{i=1}^n \lambda_i^2 $$ which implies that $$ \sum_{i=1}^n \lambda_i^2 = 1 \tag{1} $$
In addition, since $$ \int_0^1 \sum_{i=1}^n f_i^2 = n $$ we can choose $x_0 \in [0,1]$ such that $$ \sum_{i=1}^n f_i^2(x_0) \ge n \tag{2} $$
Note that $$ \left(\sum_{i=1}^n \lambda_i f_i(x_0)\right)^2 = \sum_{i=1}^n \lambda_i^2 \cdot \sum_{i=1}^n f_i^2(x_0) $$ holds if we choose appropriate $\lambda_i $.
Then via $(1)$ and $(2)$ we arrive at the conclusion.