If $I$ an interval and $x_1,x_2,x_3in I$, why $t_1x_1+t_2x_2+t_3x_3in I$ when $t_1+t_2+t_3=1, t_iin [0,1]$?
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If $I$ an interval and $x_1,x_2,x_3in I$. Why $t_1x_1+t_2x_2+t_3x_3in I$ when $t_1+t_2+t_3=1, t_iin [0,1]$ ?
Attempts : $t_3=1-t_1-t_2$ and thus $$t_1x_2+t_2x_2+t_3x_3=underbracet_1x_1+(1-t_1)x_3_in I+t_2(x_2+x_3).$$
Now, $t_1x_1+(1-t_1)x_3in I$, but this don't prove that $t_1x_1+(1-t_1)x_3+t_2(x_2+x_3)in I$. Any idea ?
real-analysis
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If $I$ an interval and $x_1,x_2,x_3in I$. Why $t_1x_1+t_2x_2+t_3x_3in I$ when $t_1+t_2+t_3=1, t_iin [0,1]$ ?
Attempts : $t_3=1-t_1-t_2$ and thus $$t_1x_2+t_2x_2+t_3x_3=underbracet_1x_1+(1-t_1)x_3_in I+t_2(x_2+x_3).$$
Now, $t_1x_1+(1-t_1)x_3in I$, but this don't prove that $t_1x_1+(1-t_1)x_3+t_2(x_2+x_3)in I$. Any idea ?
real-analysis
Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
â Kavi Rama Murthy
Aug 21 at 10:09
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up vote
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up vote
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down vote
favorite
If $I$ an interval and $x_1,x_2,x_3in I$. Why $t_1x_1+t_2x_2+t_3x_3in I$ when $t_1+t_2+t_3=1, t_iin [0,1]$ ?
Attempts : $t_3=1-t_1-t_2$ and thus $$t_1x_2+t_2x_2+t_3x_3=underbracet_1x_1+(1-t_1)x_3_in I+t_2(x_2+x_3).$$
Now, $t_1x_1+(1-t_1)x_3in I$, but this don't prove that $t_1x_1+(1-t_1)x_3+t_2(x_2+x_3)in I$. Any idea ?
real-analysis
If $I$ an interval and $x_1,x_2,x_3in I$. Why $t_1x_1+t_2x_2+t_3x_3in I$ when $t_1+t_2+t_3=1, t_iin [0,1]$ ?
Attempts : $t_3=1-t_1-t_2$ and thus $$t_1x_2+t_2x_2+t_3x_3=underbracet_1x_1+(1-t_1)x_3_in I+t_2(x_2+x_3).$$
Now, $t_1x_1+(1-t_1)x_3in I$, but this don't prove that $t_1x_1+(1-t_1)x_3+t_2(x_2+x_3)in I$. Any idea ?
real-analysis
asked Aug 21 at 10:04
Henri
1196
1196
Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
â Kavi Rama Murthy
Aug 21 at 10:09
add a comment |Â
Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
â Kavi Rama Murthy
Aug 21 at 10:09
Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
â Kavi Rama Murthy
Aug 21 at 10:09
Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
â Kavi Rama Murthy
Aug 21 at 10:09
add a comment |Â
2 Answers
2
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up vote
7
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accepted
WLOG, suppose $x_1 leqslant x_2leqslant x_3$, then
$$
x_1 =(t_1+t_2+t_3)x_1leqslant t_1 x_1 + t_2 x_2 + t_3 x_3 leqslant (t_1 + t_2 + t_3)x_3 = x_3.
$$
Since $[x_1, x_3]subset I$, so is $t_1 x_1 + t_2 x_2 + t_3x_3$.
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$$t_1x_1+t_2x_2+t_3x_3=(t_1+t_2)left(fract_1t_1+t_2x_1+fract_2t_1+t_2x_2right)+t_3x_3$$
But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
â Henri
Aug 21 at 13:09
1
The two fractions add to 1.
â Empy2
Aug 21 at 13:11
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
WLOG, suppose $x_1 leqslant x_2leqslant x_3$, then
$$
x_1 =(t_1+t_2+t_3)x_1leqslant t_1 x_1 + t_2 x_2 + t_3 x_3 leqslant (t_1 + t_2 + t_3)x_3 = x_3.
$$
Since $[x_1, x_3]subset I$, so is $t_1 x_1 + t_2 x_2 + t_3x_3$.
add a comment |Â
up vote
7
down vote
accepted
WLOG, suppose $x_1 leqslant x_2leqslant x_3$, then
$$
x_1 =(t_1+t_2+t_3)x_1leqslant t_1 x_1 + t_2 x_2 + t_3 x_3 leqslant (t_1 + t_2 + t_3)x_3 = x_3.
$$
Since $[x_1, x_3]subset I$, so is $t_1 x_1 + t_2 x_2 + t_3x_3$.
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
WLOG, suppose $x_1 leqslant x_2leqslant x_3$, then
$$
x_1 =(t_1+t_2+t_3)x_1leqslant t_1 x_1 + t_2 x_2 + t_3 x_3 leqslant (t_1 + t_2 + t_3)x_3 = x_3.
$$
Since $[x_1, x_3]subset I$, so is $t_1 x_1 + t_2 x_2 + t_3x_3$.
WLOG, suppose $x_1 leqslant x_2leqslant x_3$, then
$$
x_1 =(t_1+t_2+t_3)x_1leqslant t_1 x_1 + t_2 x_2 + t_3 x_3 leqslant (t_1 + t_2 + t_3)x_3 = x_3.
$$
Since $[x_1, x_3]subset I$, so is $t_1 x_1 + t_2 x_2 + t_3x_3$.
edited Aug 21 at 10:09
Stefan4024
29.2k53276
29.2k53276
answered Aug 21 at 10:09
xbh
2,617114
2,617114
add a comment |Â
add a comment |Â
up vote
1
down vote
$$t_1x_1+t_2x_2+t_3x_3=(t_1+t_2)left(fract_1t_1+t_2x_1+fract_2t_1+t_2x_2right)+t_3x_3$$
But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
â Henri
Aug 21 at 13:09
1
The two fractions add to 1.
â Empy2
Aug 21 at 13:11
add a comment |Â
up vote
1
down vote
$$t_1x_1+t_2x_2+t_3x_3=(t_1+t_2)left(fract_1t_1+t_2x_1+fract_2t_1+t_2x_2right)+t_3x_3$$
But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
â Henri
Aug 21 at 13:09
1
The two fractions add to 1.
â Empy2
Aug 21 at 13:11
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$t_1x_1+t_2x_2+t_3x_3=(t_1+t_2)left(fract_1t_1+t_2x_1+fract_2t_1+t_2x_2right)+t_3x_3$$
$$t_1x_1+t_2x_2+t_3x_3=(t_1+t_2)left(fract_1t_1+t_2x_1+fract_2t_1+t_2x_2right)+t_3x_3$$
answered Aug 21 at 10:52
Empy2
31.9k12059
31.9k12059
But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
â Henri
Aug 21 at 13:09
1
The two fractions add to 1.
â Empy2
Aug 21 at 13:11
add a comment |Â
But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
â Henri
Aug 21 at 13:09
1
The two fractions add to 1.
â Empy2
Aug 21 at 13:11
But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
â Henri
Aug 21 at 13:09
But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
â Henri
Aug 21 at 13:09
1
1
The two fractions add to 1.
â Empy2
Aug 21 at 13:11
The two fractions add to 1.
â Empy2
Aug 21 at 13:11
add a comment |Â
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Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
â Kavi Rama Murthy
Aug 21 at 10:09