# How to prove equivalence of two norms?

have linear span $E=Span(1,x,x^2,...,x^{2015})$ on the $C[0,1]$.

$$\left \| f \right \|_{\infty} = \underset{x\in[0,1]}{\max} \left |\sum_{i= 0}^{2015} \alpha_{i} x^{i} \right |$$ $$\left \| f \right \|_{1} = \left | \alpha_{0} \right | + ...+\left | \alpha_{2015} \right |$$

How to prove that these two norms are equivalent?

$\left \| f \right \|_{\infty} \leq \left \|f \right \|_{1}$

and

$\left \|f \right \|_{1} \leq A \left \| f \right \|_{\infty}$

First inequation is easy, but I cant prove second.

• Are you sure the sum starts from $i=1$? Dec 28, 2015 at 17:19
• sum starts from $i=0$ of course Dec 28, 2015 at 17:23

On the one hand, by triagle inequality $$\|f\|_{\infty}=|\sum_i \alpha_ix^i| \le \sum_i|\alpha_ix^i| \le \sum_i |\alpha_i|=\|f\|_1.$$ On the other hand, supposing $f\neq 0$, we have $$\frac{\|f\|_1}{\|f\|_{\infty}}=\frac{\sum_i |\alpha_i|}{\max_{x \in [0,1]} |\sum_i \alpha_ix^i|}=\frac{1}{\max_{x \in [0,1], \alpha \in S} |\sum_i \alpha_ix^i|},$$ where $S$ is the set of vectors $(\alpha_0,\ldots,\alpha_{2015})$ for which $\sum_i|\alpha_i|=1$. The denominator is a continuous functions defined on a compact set. Conclude :)
Two norms defined on a finite dimensional vector space are equivalent the standard proof is to show that $\| \|$ is equivalent to the euclidean norm $\| \|_e$. Consider $f:R^n\rightarrow R$ defined by $f(x)=\|x\|$ its restriction on the the unit euclidean ball reach its $inf =m$ and its $max=M$ since the unit euclidean ball is compact so:
$m\leq \|x/\|x\|_e\|\leq M$ thus $m\|x\|_e\leq \|x\|\leq M\|x\|_e$
$$\| f \|_1 = |a_0| + |a_1| + \dots + |a_{2015}| \le 2016 \max \limits _{0 \le i \le 2015} |a_i| \le 2016 \| f \|_\infty \implies A = 2016$$