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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.

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    $\begingroup$ Are you sure the sum starts from $i=1$? $\endgroup$ Dec 28, 2015 at 17:19
  • $\begingroup$ sum starts from $i=0$ of course $\endgroup$
    – Roman F.
    Dec 28, 2015 at 17:23

3 Answers 3

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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 :)

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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 $m\|x\|_e\leq \|x\|\leq M\|x\|_e$

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  • $\begingroup$ but it took a span which is finite dimensional $\endgroup$ Dec 28, 2015 at 17:33
  • $\begingroup$ I see your point :) $\endgroup$ Dec 28, 2015 at 17:36
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$$\| f \|_1 = |a_0| + |a_1| + \dots + |a_{2015}| \le 2016 \max \limits _{0 \le i \le 2015} |a_i| \le 2016 \| f \|_\infty \implies A = 2016$$

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