Show $sup_{xin[0,1]} | x -f(x)|geq frac{1}{2}$ for any $f$ with $f(0)=0$ and $int^1_0 f(t)dt=0$. [closed]
$begingroup$
Let $f:[0,1]to mathbb R$ be a continuous function with $f(0)=0$ and $int^1_0 f(t)dt=0$.
I need to show that $$sup_{xin[0,1]} | x -f(x)|geq frac{1}{2}$$
I'm not sure how to approach this problem. Any tips are greatly appreciated!
real-analysis inequality
edited Dec 3 '18 at 18:18
José Carlos Santos
163k22131234
asked Dec 3 '18 at 18:12
user3342072user3342072
400213
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closed as off-topic by greedoid, Alexander Gruber♦ Dec 4 '18 at 4:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – greedoid, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
Let $f:[0,1]to mathbb R$ be a continuous function with $f(0)=0$ and $int^1_0 f(t)dt=0$.
I need to show that $$sup_{xin[0,1]} | x -f(x)|geq frac{1}{2}$$
I'm not sure how to approach this problem. Any tips are greatly appreciated!
real-analysis inequality
edited Dec 3 '18 at 18:18
José Carlos Santos
163k22131234
asked Dec 3 '18 at 18:12
user3342072user3342072
400213
$endgroup$
closed as off-topic by greedoid, Alexander Gruber♦ Dec 4 '18 at 4:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – greedoid, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $f:[0,1]to mathbb R$ be a continuous function with $f(0)=0$ and $int^1_0 f(t)dt=0$.
I need to show that $$sup_{xin[0,1]} | x -f(x)|geq frac{1}{2}$$
I'm not sure how to approach this problem. Any tips are greatly appreciated!
real-analysis inequality
edited Dec 3 '18 at 18:18
José Carlos Santos
163k22131234
asked Dec 3 '18 at 18:12
user3342072user3342072
400213
$endgroup$
Let $f:[0,1]to mathbb R$ be a continuous function with $f(0)=0$ and $int^1_0 f(t)dt=0$.
I need to show that $$sup_{xin[0,1]} | x -f(x)|geq frac{1}{2}$$
I'm not sure how to approach this problem. Any tips are greatly appreciated!
real-analysis inequality
real-analysis inequality
edited Dec 3 '18 at 18:18
José Carlos Santos
163k22131234
asked Dec 3 '18 at 18:12
user3342072user3342072
400213
edited Dec 3 '18 at 18:18
José Carlos Santos
163k22131234
asked Dec 3 '18 at 18:12
user3342072user3342072
400213
edited Dec 3 '18 at 18:18
José Carlos Santos
163k22131234
edited Dec 3 '18 at 18:18
José Carlos Santos
163k22131234
edited Dec 3 '18 at 18:18
José Carlos Santos
163k22131234
163k22131234
asked Dec 3 '18 at 18:12
user3342072user3342072
400213
asked Dec 3 '18 at 18:12
user3342072user3342072
400213
asked Dec 3 '18 at 18:12
user3342072user3342072
400213
400213
closed as off-topic by greedoid, Alexander Gruber♦ Dec 4 '18 at 4:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – greedoid, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by greedoid, Alexander Gruber♦ Dec 4 '18 at 4:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – greedoid, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
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Hint: $displaystyleint_0^1x-f(x),mathrm dx=frac12$
answered Dec 3 '18 at 18:16
José Carlos SantosJosé Carlos Santos
163k22131234
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Suppose instead that $lvert x - f(x) rvert < 1/2$ for all $x in [0,1]$. Then we see that $$frac 1 2 = left lvert int^1_0 (x-f(x)) dxright rvert le int^1_0 lvert x - f(x)rvert dx < int^1_0 frac 1 2 dx = frac 1 2,$$ a contradiction. Thus we must have $lvert x - f(x) rvert ge 1/2$ for some $x in [0,1]$ and the conclusion follows.
Unless I've made a mistake, it seems the assumption that $f(0) = 0$ is unnecessary.
answered Dec 3 '18 at 18:16
User8128User8128
10.8k1622
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $displaystyleint_0^1x-f(x),mathrm dx=frac12$
answered Dec 3 '18 at 18:16
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
Hint: $displaystyleint_0^1x-f(x),mathrm dx=frac12$
answered Dec 3 '18 at 18:16
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
Hint: $displaystyleint_0^1x-f(x),mathrm dx=frac12$
answered Dec 3 '18 at 18:16
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
Hint: $displaystyleint_0^1x-f(x),mathrm dx=frac12$
answered Dec 3 '18 at 18:16
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 3 '18 at 18:16
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 3 '18 at 18:16
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 3 '18 at 18:16
José Carlos SantosJosé Carlos Santos
163k22131234
163k22131234
add a comment |
add a comment |
$begingroup$
Suppose instead that $lvert x - f(x) rvert < 1/2$ for all $x in [0,1]$. Then we see that $$frac 1 2 = left lvert int^1_0 (x-f(x)) dxright rvert le int^1_0 lvert x - f(x)rvert dx < int^1_0 frac 1 2 dx = frac 1 2,$$ a contradiction. Thus we must have $lvert x - f(x) rvert ge 1/2$ for some $x in [0,1]$ and the conclusion follows.
Unless I've made a mistake, it seems the assumption that $f(0) = 0$ is unnecessary.
answered Dec 3 '18 at 18:16
User8128User8128
10.8k1622
$endgroup$
add a comment |
$begingroup$
Suppose instead that $lvert x - f(x) rvert < 1/2$ for all $x in [0,1]$. Then we see that $$frac 1 2 = left lvert int^1_0 (x-f(x)) dxright rvert le int^1_0 lvert x - f(x)rvert dx < int^1_0 frac 1 2 dx = frac 1 2,$$ a contradiction. Thus we must have $lvert x - f(x) rvert ge 1/2$ for some $x in [0,1]$ and the conclusion follows.
Unless I've made a mistake, it seems the assumption that $f(0) = 0$ is unnecessary.
answered Dec 3 '18 at 18:16
User8128User8128
10.8k1622
$endgroup$
add a comment |
$begingroup$
Suppose instead that $lvert x - f(x) rvert < 1/2$ for all $x in [0,1]$. Then we see that $$frac 1 2 = left lvert int^1_0 (x-f(x)) dxright rvert le int^1_0 lvert x - f(x)rvert dx < int^1_0 frac 1 2 dx = frac 1 2,$$ a contradiction. Thus we must have $lvert x - f(x) rvert ge 1/2$ for some $x in [0,1]$ and the conclusion follows.
Unless I've made a mistake, it seems the assumption that $f(0) = 0$ is unnecessary.
answered Dec 3 '18 at 18:16
User8128User8128
10.8k1622
$endgroup$
Suppose instead that $lvert x - f(x) rvert < 1/2$ for all $x in [0,1]$. Then we see that $$frac 1 2 = left lvert int^1_0 (x-f(x)) dxright rvert le int^1_0 lvert x - f(x)rvert dx < int^1_0 frac 1 2 dx = frac 1 2,$$ a contradiction. Thus we must have $lvert x - f(x) rvert ge 1/2$ for some $x in [0,1]$ and the conclusion follows.
Unless I've made a mistake, it seems the assumption that $f(0) = 0$ is unnecessary.
answered Dec 3 '18 at 18:16
User8128User8128
10.8k1622
answered Dec 3 '18 at 18:16
User8128User8128
10.8k1622
answered Dec 3 '18 at 18:16
User8128User8128
10.8k1622
answered Dec 3 '18 at 18:16
User8128User8128
10.8k1622
10.8k1622
add a comment |
add a comment |
