Show that $sum_{k=0}^{N} |P(k)| leq C(N) int_{0} ^{1} |P(t)| dt $.
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
functional-analysis polynomials norm normed-spaces upper-lower-bounds
edited Dec 9 at 8:14
Batominovski
33.7k33292
asked Dec 8 at 23:33
Maggie94
374
add a comment |
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
functional-analysis polynomials norm normed-spaces upper-lower-bounds
edited Dec 9 at 8:14
Batominovski
33.7k33292
asked Dec 8 at 23:33
Maggie94
374
add a comment |
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
functional-analysis polynomials norm normed-spaces upper-lower-bounds
edited Dec 9 at 8:14
Batominovski
33.7k33292
asked Dec 8 at 23:33
Maggie94
374
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
functional-analysis polynomials norm normed-spaces upper-lower-bounds
functional-analysis polynomials norm normed-spaces upper-lower-bounds
edited Dec 9 at 8:14
Batominovski
33.7k33292
asked Dec 8 at 23:33
Maggie94
374
edited Dec 9 at 8:14
Batominovski
33.7k33292
asked Dec 8 at 23:33
Maggie94
374
edited Dec 9 at 8:14
Batominovski
33.7k33292
edited Dec 9 at 8:14
Batominovski
33.7k33292
edited Dec 9 at 8:14
Batominovski
33.7k33292
33.7k33292
asked Dec 8 at 23:33
Maggie94
374
asked Dec 8 at 23:33
Maggie94
374
asked Dec 8 at 23:33
Maggie94
374
374
add a comment |
add a comment |
1 Answer
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Let $V_N$ be the complex vector space of polynomials in $mathbb{C}[t]$ of degree at most $Ninmathbb{Z}_{geq 0}$. 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 $Pin V_N$ 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 $ainmathbb{C}$ and $Pin V_N$. If $|P|_1=0$ for some $Pin V_N$, then $P(0)=P(1)=P(2)=ldots=P(N)=0$, so $$P(t)=Q(t),prod_{k=0}^N,(t-k)$$ for some $Q(t)inmathbb{C}[t]$. As $P$ has degree at most $N$, $Q$ must be identically $0$, so $Pequiv 0$. If $P_1,P_2in V_N$, then $P_1+P_2in V_N$, and $$(P_1+P_2)(k)=P_1(k)+P_2(k)text{ for every }kinmathbb{C},,$$ whence $$big|(P_1+P_2)(k)big|=big|P_1(k)+P_2(k)big|leq big|P_1(k)big|+big|P_2(k)big|text{ for every }kinmathbb{C},.$$ Taking the sum over $k=0,1,2,ldots,N$, we get $|P_1+P_2|_1leq |P_1|_1+|P_2|_1$, 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 $c(N),C(N)inmathbb{R}_{>0}$ such that
$$c(N),|P|_2leq |P|_1leq C(N),|P|_2$$
for all $Pin V_N$.
answered Dec 8 at 23:54
Batominovski
33.7k33292
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 at 13:03
Yes, you are right.
– Batominovski
Dec 10 at 20:03
add a comment |
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1 Answer
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1 Answer
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Let $V_N$ be the complex vector space of polynomials in $mathbb{C}[t]$ of degree at most $Ninmathbb{Z}_{geq 0}$. 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 $Pin V_N$ 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 $ainmathbb{C}$ and $Pin V_N$. If $|P|_1=0$ for some $Pin V_N$, then $P(0)=P(1)=P(2)=ldots=P(N)=0$, so $$P(t)=Q(t),prod_{k=0}^N,(t-k)$$ for some $Q(t)inmathbb{C}[t]$. As $P$ has degree at most $N$, $Q$ must be identically $0$, so $Pequiv 0$. If $P_1,P_2in V_N$, then $P_1+P_2in V_N$, and $$(P_1+P_2)(k)=P_1(k)+P_2(k)text{ for every }kinmathbb{C},,$$ whence $$big|(P_1+P_2)(k)big|=big|P_1(k)+P_2(k)big|leq big|P_1(k)big|+big|P_2(k)big|text{ for every }kinmathbb{C},.$$ Taking the sum over $k=0,1,2,ldots,N$, we get $|P_1+P_2|_1leq |P_1|_1+|P_2|_1$, 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 $c(N),C(N)inmathbb{R}_{>0}$ such that
$$c(N),|P|_2leq |P|_1leq C(N),|P|_2$$
for all $Pin V_N$.
answered Dec 8 at 23:54
Batominovski
33.7k33292
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 at 13:03
Yes, you are right.
– Batominovski
Dec 10 at 20:03
add a comment |
Let $V_N$ be the complex vector space of polynomials in $mathbb{C}[t]$ of degree at most $Ninmathbb{Z}_{geq 0}$. 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 $Pin V_N$ 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 $ainmathbb{C}$ and $Pin V_N$. If $|P|_1=0$ for some $Pin V_N$, then $P(0)=P(1)=P(2)=ldots=P(N)=0$, so $$P(t)=Q(t),prod_{k=0}^N,(t-k)$$ for some $Q(t)inmathbb{C}[t]$. As $P$ has degree at most $N$, $Q$ must be identically $0$, so $Pequiv 0$. If $P_1,P_2in V_N$, then $P_1+P_2in V_N$, and $$(P_1+P_2)(k)=P_1(k)+P_2(k)text{ for every }kinmathbb{C},,$$ whence $$big|(P_1+P_2)(k)big|=big|P_1(k)+P_2(k)big|leq big|P_1(k)big|+big|P_2(k)big|text{ for every }kinmathbb{C},.$$ Taking the sum over $k=0,1,2,ldots,N$, we get $|P_1+P_2|_1leq |P_1|_1+|P_2|_1$, 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 $c(N),C(N)inmathbb{R}_{>0}$ such that
$$c(N),|P|_2leq |P|_1leq C(N),|P|_2$$
for all $Pin V_N$.
answered Dec 8 at 23:54
Batominovski
33.7k33292
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 at 13:03
Yes, you are right.
– Batominovski
Dec 10 at 20:03
add a comment |
Let $V_N$ be the complex vector space of polynomials in $mathbb{C}[t]$ of degree at most $Ninmathbb{Z}_{geq 0}$. 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 $Pin V_N$ 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 $ainmathbb{C}$ and $Pin V_N$. If $|P|_1=0$ for some $Pin V_N$, then $P(0)=P(1)=P(2)=ldots=P(N)=0$, so $$P(t)=Q(t),prod_{k=0}^N,(t-k)$$ for some $Q(t)inmathbb{C}[t]$. As $P$ has degree at most $N$, $Q$ must be identically $0$, so $Pequiv 0$. If $P_1,P_2in V_N$, then $P_1+P_2in V_N$, and $$(P_1+P_2)(k)=P_1(k)+P_2(k)text{ for every }kinmathbb{C},,$$ whence $$big|(P_1+P_2)(k)big|=big|P_1(k)+P_2(k)big|leq big|P_1(k)big|+big|P_2(k)big|text{ for every }kinmathbb{C},.$$ Taking the sum over $k=0,1,2,ldots,N$, we get $|P_1+P_2|_1leq |P_1|_1+|P_2|_1$, 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 $c(N),C(N)inmathbb{R}_{>0}$ such that
$$c(N),|P|_2leq |P|_1leq C(N),|P|_2$$
for all $Pin V_N$.
answered Dec 8 at 23:54
Batominovski
33.7k33292
Let $V_N$ be the complex vector space of polynomials in $mathbb{C}[t]$ of degree at most $Ninmathbb{Z}_{geq 0}$. 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 $Pin V_N$ 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 $ainmathbb{C}$ and $Pin V_N$. If $|P|_1=0$ for some $Pin V_N$, then $P(0)=P(1)=P(2)=ldots=P(N)=0$, so $$P(t)=Q(t),prod_{k=0}^N,(t-k)$$ for some $Q(t)inmathbb{C}[t]$. As $P$ has degree at most $N$, $Q$ must be identically $0$, so $Pequiv 0$. If $P_1,P_2in V_N$, then $P_1+P_2in V_N$, and $$(P_1+P_2)(k)=P_1(k)+P_2(k)text{ for every }kinmathbb{C},,$$ whence $$big|(P_1+P_2)(k)big|=big|P_1(k)+P_2(k)big|leq big|P_1(k)big|+big|P_2(k)big|text{ for every }kinmathbb{C},.$$ Taking the sum over $k=0,1,2,ldots,N$, we get $|P_1+P_2|_1leq |P_1|_1+|P_2|_1$, 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 $c(N),C(N)inmathbb{R}_{>0}$ such that
$$c(N),|P|_2leq |P|_1leq C(N),|P|_2$$
for all $Pin V_N$.
answered Dec 8 at 23:54
Batominovski
33.7k33292
edited Dec 9 at 8:08
answered Dec 8 at 23:54
Batominovski
33.7k33292
answered Dec 8 at 23:54
Batominovski
33.7k33292
answered Dec 8 at 23:54
Batominovski
33.7k33292
33.7k33292
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 at 13:03
Yes, you are right.
– Batominovski
Dec 10 at 20:03
add a comment |
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 at 13:03
Yes, you are right.
– Batominovski
Dec 10 at 20:03
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 at 13:03
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 at 13:03
Yes, you are right.
– Batominovski
Dec 10 at 20:03
Yes, you are right.
– Batominovski
Dec 10 at 20:03
add a comment |
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