Mixing
$L^1$ inequality between ordered real numbers implies $L^2$ norm inequality
Let $x_1,ldots,x_n$ be real numbers such that $|x_1|geq ldotsgeq |x_n|$ and $displaystyle sum_{i=k+1}^n |x_i| leq alpha sum_{i=1}^k |x_i|$ where $1leq k leq n-1$ and $alpha >0$.
Prove that $$sum_{i=k+1}^n x_i^2 leq alpha sum_{i=1}^k x_i^2$$
I have managed to prove a looser inequality: $$begin{align}
alpha sum_{i=1}^k x_i^2
&geq frac{alpha}{k}left(sum_{i=1}^k |x_i| right)^2 quad quad text{Cauchy-Schwarz}\
&geq frac 1k sum_{i=k+1}^n |x_i| sum_{i=1}^k |x_i|\
&geq frac 1k left(sum_{i=k+1}^n |x_i| right)^2\
&geq frac 1k sum_{i=k+1}^n x_i^2 quad quad text{since } |z|_1 geq |z|_2
end{align}$$
Since $frac 1k$ is not $geq 1$, I'm stuck...
real-analysis inequality cauchy-schwarz-inequality
edited Nov 21 '18 at 11:28
Tianlalu
3,09621038
asked Nov 21 '18 at 10:15
Issou Chankla
254
add a comment |
Let $x_1,ldots,x_n$ be real numbers such that $|x_1|geq ldotsgeq |x_n|$ and $displaystyle sum_{i=k+1}^n |x_i| leq alpha sum_{i=1}^k |x_i|$ where $1leq k leq n-1$ and $alpha >0$.
Prove that $$sum_{i=k+1}^n x_i^2 leq alpha sum_{i=1}^k x_i^2$$
I have managed to prove a looser inequality: $$begin{align}
alpha sum_{i=1}^k x_i^2
&geq frac{alpha}{k}left(sum_{i=1}^k |x_i| right)^2 quad quad text{Cauchy-Schwarz}\
&geq frac 1k sum_{i=k+1}^n |x_i| sum_{i=1}^k |x_i|\
&geq frac 1k left(sum_{i=k+1}^n |x_i| right)^2\
&geq frac 1k sum_{i=k+1}^n x_i^2 quad quad text{since } |z|_1 geq |z|_2
end{align}$$
Since $frac 1k$ is not $geq 1$, I'm stuck...
real-analysis inequality cauchy-schwarz-inequality
edited Nov 21 '18 at 11:28
Tianlalu
3,09621038
asked Nov 21 '18 at 10:15
Issou Chankla
254
add a comment |
Let $x_1,ldots,x_n$ be real numbers such that $|x_1|geq ldotsgeq |x_n|$ and $displaystyle sum_{i=k+1}^n |x_i| leq alpha sum_{i=1}^k |x_i|$ where $1leq k leq n-1$ and $alpha >0$.
Prove that $$sum_{i=k+1}^n x_i^2 leq alpha sum_{i=1}^k x_i^2$$
I have managed to prove a looser inequality: $$begin{align}
alpha sum_{i=1}^k x_i^2
&geq frac{alpha}{k}left(sum_{i=1}^k |x_i| right)^2 quad quad text{Cauchy-Schwarz}\
&geq frac 1k sum_{i=k+1}^n |x_i| sum_{i=1}^k |x_i|\
&geq frac 1k left(sum_{i=k+1}^n |x_i| right)^2\
&geq frac 1k sum_{i=k+1}^n x_i^2 quad quad text{since } |z|_1 geq |z|_2
end{align}$$
Since $frac 1k$ is not $geq 1$, I'm stuck...
real-analysis inequality cauchy-schwarz-inequality
edited Nov 21 '18 at 11:28
Tianlalu
3,09621038
asked Nov 21 '18 at 10:15
Issou Chankla
254
Let $x_1,ldots,x_n$ be real numbers such that $|x_1|geq ldotsgeq |x_n|$ and $displaystyle sum_{i=k+1}^n |x_i| leq alpha sum_{i=1}^k |x_i|$ where $1leq k leq n-1$ and $alpha >0$.
Prove that $$sum_{i=k+1}^n x_i^2 leq alpha sum_{i=1}^k x_i^2$$
I have managed to prove a looser inequality: $$begin{align}
alpha sum_{i=1}^k x_i^2
&geq frac{alpha}{k}left(sum_{i=1}^k |x_i| right)^2 quad quad text{Cauchy-Schwarz}\
&geq frac 1k sum_{i=k+1}^n |x_i| sum_{i=1}^k |x_i|\
&geq frac 1k left(sum_{i=k+1}^n |x_i| right)^2\
&geq frac 1k sum_{i=k+1}^n x_i^2 quad quad text{since } |z|_1 geq |z|_2
end{align}$$
Since $frac 1k$ is not $geq 1$, I'm stuck...
real-analysis inequality cauchy-schwarz-inequality
real-analysis inequality cauchy-schwarz-inequality
edited Nov 21 '18 at 11:28
Tianlalu
3,09621038
asked Nov 21 '18 at 10:15
Issou Chankla
254
edited Nov 21 '18 at 11:28
Tianlalu
3,09621038
asked Nov 21 '18 at 10:15
Issou Chankla
254
edited Nov 21 '18 at 11:28
Tianlalu
3,09621038
edited Nov 21 '18 at 11:28
Tianlalu
3,09621038
edited Nov 21 '18 at 11:28
Tianlalu
3,09621038
3,09621038
asked Nov 21 '18 at 10:15
Issou Chankla
254
asked Nov 21 '18 at 10:15
Issou Chankla
254
asked Nov 21 '18 at 10:15
Issou Chankla
254
254
add a comment |
add a comment |
1 Answer
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It is quite simple: Note that $|x_i| le |x_k|$ for any $i in {k+1,ldots,n}$. Thus
$$tag{1}sum_{i=k+1}^n x_i^2 le |x_k| sum_{i=k+1}^n |x_i| le alpha |x_k| sum_{i=1}^k |x_i|,$$
where the given condition is used in the last step. Now $|x_k x_i| le x_i^2$ for an $i=1,ldots,k$, because $|x_k| le |x_i|$ for all $i=1,ldots,k$. So proceeding in (1) gives the claimed estimate
$$sum_{i=k+1}^n x_i^2lealpha sum_{i=1}^k x_i^2.$$
answered Nov 21 '18 at 11:06
p4sch
4,770217
Nice, thank you. This line of thought did not occur to me.
– Issou Chankla
Nov 21 '18 at 11:17
In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
– p4sch
Nov 21 '18 at 12:04
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
It is quite simple: Note that $|x_i| le |x_k|$ for any $i in {k+1,ldots,n}$. Thus
$$tag{1}sum_{i=k+1}^n x_i^2 le |x_k| sum_{i=k+1}^n |x_i| le alpha |x_k| sum_{i=1}^k |x_i|,$$
where the given condition is used in the last step. Now $|x_k x_i| le x_i^2$ for an $i=1,ldots,k$, because $|x_k| le |x_i|$ for all $i=1,ldots,k$. So proceeding in (1) gives the claimed estimate
$$sum_{i=k+1}^n x_i^2lealpha sum_{i=1}^k x_i^2.$$
answered Nov 21 '18 at 11:06
p4sch
4,770217
Nice, thank you. This line of thought did not occur to me.
– Issou Chankla
Nov 21 '18 at 11:17
In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
– p4sch
Nov 21 '18 at 12:04
add a comment |
It is quite simple: Note that $|x_i| le |x_k|$ for any $i in {k+1,ldots,n}$. Thus
$$tag{1}sum_{i=k+1}^n x_i^2 le |x_k| sum_{i=k+1}^n |x_i| le alpha |x_k| sum_{i=1}^k |x_i|,$$
where the given condition is used in the last step. Now $|x_k x_i| le x_i^2$ for an $i=1,ldots,k$, because $|x_k| le |x_i|$ for all $i=1,ldots,k$. So proceeding in (1) gives the claimed estimate
$$sum_{i=k+1}^n x_i^2lealpha sum_{i=1}^k x_i^2.$$
answered Nov 21 '18 at 11:06
p4sch
4,770217
Nice, thank you. This line of thought did not occur to me.
– Issou Chankla
Nov 21 '18 at 11:17
In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
– p4sch
Nov 21 '18 at 12:04
add a comment |
It is quite simple: Note that $|x_i| le |x_k|$ for any $i in {k+1,ldots,n}$. Thus
$$tag{1}sum_{i=k+1}^n x_i^2 le |x_k| sum_{i=k+1}^n |x_i| le alpha |x_k| sum_{i=1}^k |x_i|,$$
where the given condition is used in the last step. Now $|x_k x_i| le x_i^2$ for an $i=1,ldots,k$, because $|x_k| le |x_i|$ for all $i=1,ldots,k$. So proceeding in (1) gives the claimed estimate
$$sum_{i=k+1}^n x_i^2lealpha sum_{i=1}^k x_i^2.$$
answered Nov 21 '18 at 11:06
p4sch
4,770217
It is quite simple: Note that $|x_i| le |x_k|$ for any $i in {k+1,ldots,n}$. Thus
$$tag{1}sum_{i=k+1}^n x_i^2 le |x_k| sum_{i=k+1}^n |x_i| le alpha |x_k| sum_{i=1}^k |x_i|,$$
where the given condition is used in the last step. Now $|x_k x_i| le x_i^2$ for an $i=1,ldots,k$, because $|x_k| le |x_i|$ for all $i=1,ldots,k$. So proceeding in (1) gives the claimed estimate
$$sum_{i=k+1}^n x_i^2lealpha sum_{i=1}^k x_i^2.$$
answered Nov 21 '18 at 11:06
p4sch
4,770217
answered Nov 21 '18 at 11:06
p4sch
4,770217
answered Nov 21 '18 at 11:06
p4sch
4,770217
answered Nov 21 '18 at 11:06
p4sch
4,770217
4,770217
Nice, thank you. This line of thought did not occur to me.
– Issou Chankla
Nov 21 '18 at 11:17
In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
– p4sch
Nov 21 '18 at 12:04
add a comment |
Nice, thank you. This line of thought did not occur to me.
– Issou Chankla
Nov 21 '18 at 11:17
In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
– p4sch
Nov 21 '18 at 12:04
Nice, thank you. This line of thought did not occur to me.
– Issou Chankla
Nov 21 '18 at 11:17
Nice, thank you. This line of thought did not occur to me.
– Issou Chankla
Nov 21 '18 at 11:17
In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
– p4sch
Nov 21 '18 at 12:04
In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
– p4sch
Nov 21 '18 at 12:04
add a comment |
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