# Show that $\sum_{k=0}^{N} |P(k)| \leq C(N) \int_{0} ^{1} |P(t)| dt$. [closed]

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

## closed as off-topic by user21820, RRL, Did, Xander Henderson, José Carlos SantosFeb 14 at 14:11

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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 $$P\in V_N$$.
• 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? – Maggie94 Dec 10 '18 at 13:03