I’m attending a functional analysis course and I am given to solve this problem as an exercise but I’m a little bit disoriented and I don’t know what tools I can use to get it.

Show that, for each $N \in \mathbb{N}$ , there exists a constant $C(N) \in \mathbb{R}^+$ such that if $P$ in a polynomial of degree $N$ with complex coefficients, we have $$\sum_{k=0}^{N} |P(k)| \leq C(N) \int_{0} ^{1} |P(t)| dt\,.$$

Any idea or hint? Thank you all in advance


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Let $V_N$ be the complex vector space of polynomials in $\mathbb{C}[t]$ of degree at most $N\in\mathbb{Z}_{\geq 0}$. Show that $\|\_\|_1$ and $\|\_\|_2$ defined by $$\|P\|_1:=\sum_{k=0}^N\,\big|P(k)\big|$$ and $$\|P\|_2:=\int_0^1\,\big|P(t)\big|\,\text{d}t$$ for all $P\in V_N$ are norms on $V_N$.

To show that $\|\_\|_1$ is a norm, we note that $\|a\, P\|_1=|a|\,\|P\|_1$ trivially holds for all $a\in\mathbb{C}$ and $P\in V_N$. If $\|P\|_1=0$ for some $P\in V_N$, then $P(0)=P(1)=P(2)=\ldots=P(N)=0$, so $$P(t)=Q(t)\,\prod_{k=0}^N\,(t-k)$$ for some $Q(t)\in\mathbb{C}[t]$. As $P$ has degree at most $N$, $Q$ must be identically $0$, so $P\equiv 0$. If $P_1,P_2\in V_N$, then $P_1+P_2\in V_N$, and $$(P_1+P_2)(k)=P_1(k)+P_2(k)\text{ for every }k\in\mathbb{C}\,,$$ whence $$\big|(P_1+P_2)(k)\big|=\big|P_1(k)+P_2(k)\big|\leq \big|P_1(k)\big|+\big|P_2(k)\big|\text{ for every }k\in\mathbb{C}\,.$$ Taking the sum over $k=0,1,2,\ldots,N$, we get $\|P_1+P_2\|_1\leq \|P_1\|_1+\|P_2\|_1$, justifying the triangle inequality condition. Similarly, $\|\_\|_2$ is a norm on $V_N$.

Note that all norms on a finite-dimensional complex vector space are equivalent. That means, there exist $c(N),C(N)\in\mathbb{R}_{>0}$ such that $$c(N)\,\|P\|_2\leq \|P\|_1\leq C(N)\,\|P\|_2$$ for all $P\in V_N$.

  • $\begingroup$ Thank you! I can observe that the polynomial space I am considering has $N+1$ as dimension because it is generated, for example, by $\{x^{N},x^{N-1},...,x,1\}$ that has dimension $N+1$, which is finite. Right? $\endgroup$ – Maggie94 Dec 10 '18 at 13:03
  • $\begingroup$ Yes, you are right. $\endgroup$ – Batominovski Dec 10 '18 at 20:03

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