How to prove the following by Cauchy-Schwarz? [duplicate]
$begingroup$
This question already has an answer here:
Let $f:[0;1]tomathbb R$ be continiously differentiable [closed]
2 answers
If
$u(x) in C([a, b]), u(a) = 0,; u(x) = int_{a}^{x}u^{'}(t)dt$
then
$int_{a}^{b} |u|^{2} dx le frac{1}{2}(b - a)^{2}int_{a}^{b}|u^{'}(t)|^{2}dt$
The book said it can be proved using cauchy-schwarz-inequality, but I cannot make it.
inequality cauchy-schwarz-inequality
edited Dec 22 '18 at 17:10
J.G.
25.6k22539
asked Dec 22 '18 at 16:46
BluedropsBluedrops
136
$endgroup$
marked as duplicate by Martin R, mrtaurho, idm, Macavity inequality
Users with the inequality badge can single-handedly close inequality questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 22 '18 at 18:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Let $f:[0;1]tomathbb R$ be continiously differentiable [closed]
2 answers
If
$u(x) in C([a, b]), u(a) = 0,; u(x) = int_{a}^{x}u^{'}(t)dt$
then
$int_{a}^{b} |u|^{2} dx le frac{1}{2}(b - a)^{2}int_{a}^{b}|u^{'}(t)|^{2}dt$
The book said it can be proved using cauchy-schwarz-inequality, but I cannot make it.
inequality cauchy-schwarz-inequality
edited Dec 22 '18 at 17:10
J.G.
25.6k22539
asked Dec 22 '18 at 16:46
BluedropsBluedrops
136
$endgroup$
marked as duplicate by Martin R, mrtaurho, idm, Macavity inequality
Users with the inequality badge can single-handedly close inequality questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 22 '18 at 18:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 16:47
$begingroup$
There is no conclusion after ‘then’, only a formula.
$endgroup$
– Bernard
Dec 22 '18 at 16:52
add a comment |
$begingroup$
This question already has an answer here:
Let $f:[0;1]tomathbb R$ be continiously differentiable [closed]
2 answers
If
$u(x) in C([a, b]), u(a) = 0,; u(x) = int_{a}^{x}u^{'}(t)dt$
then
$int_{a}^{b} |u|^{2} dx le frac{1}{2}(b - a)^{2}int_{a}^{b}|u^{'}(t)|^{2}dt$
The book said it can be proved using cauchy-schwarz-inequality, but I cannot make it.
inequality cauchy-schwarz-inequality
edited Dec 22 '18 at 17:10
J.G.
25.6k22539
asked Dec 22 '18 at 16:46
BluedropsBluedrops
136
$endgroup$
This question already has an answer here:
Let $f:[0;1]tomathbb R$ be continiously differentiable [closed]
2 answers
If
$u(x) in C([a, b]), u(a) = 0,; u(x) = int_{a}^{x}u^{'}(t)dt$
then
$int_{a}^{b} |u|^{2} dx le frac{1}{2}(b - a)^{2}int_{a}^{b}|u^{'}(t)|^{2}dt$
The book said it can be proved using cauchy-schwarz-inequality, but I cannot make it.
This question already has an answer here:
Let $f:[0;1]tomathbb R$ be continiously differentiable [closed]
2 answers
inequality cauchy-schwarz-inequality
inequality cauchy-schwarz-inequality
edited Dec 22 '18 at 17:10
J.G.
25.6k22539
asked Dec 22 '18 at 16:46
BluedropsBluedrops
136
edited Dec 22 '18 at 17:10
J.G.
25.6k22539
asked Dec 22 '18 at 16:46
BluedropsBluedrops
136
edited Dec 22 '18 at 17:10
J.G.
25.6k22539
edited Dec 22 '18 at 17:10
J.G.
25.6k22539
edited Dec 22 '18 at 17:10
J.G.
25.6k22539
25.6k22539
asked Dec 22 '18 at 16:46
BluedropsBluedrops
136
asked Dec 22 '18 at 16:46
BluedropsBluedrops
136
asked Dec 22 '18 at 16:46
BluedropsBluedrops
136
136
marked as duplicate by Martin R, mrtaurho, idm, Macavity inequality
Users with the inequality badge can single-handedly close inequality questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 22 '18 at 18:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, mrtaurho, idm, Macavity inequality
Users with the inequality badge can single-handedly close inequality questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Dec 22 '18 at 18:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 16:47
$begingroup$
There is no conclusion after ‘then’, only a formula.
$endgroup$
– Bernard
Dec 22 '18 at 16:52
add a comment |
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 16:47
$begingroup$
There is no conclusion after ‘then’, only a formula.
$endgroup$
– Bernard
Dec 22 '18 at 16:52
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 16:47
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 16:47
$begingroup$
There is no conclusion after ‘then’, only a formula.
$endgroup$
– Bernard
Dec 22 '18 at 16:52
$begingroup$
There is no conclusion after ‘then’, only a formula.
$endgroup$
– Bernard
Dec 22 '18 at 16:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One long line proves it: $$int_a^b |u(t)|^2 dt=int_a^bleft|int_a^t 1cdot u'(t')dt'right|^2 dtleint_a^bleft[int_a^t 1^2 dt'cdotint_a^t |u'(t')|^2dt'right] dt\leint_a^bleft[(t-a)cdotint_a^b |u'(t')|^2dt'right] dt=frac{(b-a)^2}{2}int_a^b |u'(t')|^2dt'.$$The first $le$ uses Cauchy-Schwarz; the second replaces an $int_a^t dt'$ with $int_a^b dt'$.
answered Dec 22 '18 at 17:06
J.G.J.G.
25.6k22539
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One long line proves it: $$int_a^b |u(t)|^2 dt=int_a^bleft|int_a^t 1cdot u'(t')dt'right|^2 dtleint_a^bleft[int_a^t 1^2 dt'cdotint_a^t |u'(t')|^2dt'right] dt\leint_a^bleft[(t-a)cdotint_a^b |u'(t')|^2dt'right] dt=frac{(b-a)^2}{2}int_a^b |u'(t')|^2dt'.$$The first $le$ uses Cauchy-Schwarz; the second replaces an $int_a^t dt'$ with $int_a^b dt'$.
answered Dec 22 '18 at 17:06
J.G.J.G.
25.6k22539
$endgroup$
add a comment |
$begingroup$
One long line proves it: $$int_a^b |u(t)|^2 dt=int_a^bleft|int_a^t 1cdot u'(t')dt'right|^2 dtleint_a^bleft[int_a^t 1^2 dt'cdotint_a^t |u'(t')|^2dt'right] dt\leint_a^bleft[(t-a)cdotint_a^b |u'(t')|^2dt'right] dt=frac{(b-a)^2}{2}int_a^b |u'(t')|^2dt'.$$The first $le$ uses Cauchy-Schwarz; the second replaces an $int_a^t dt'$ with $int_a^b dt'$.
answered Dec 22 '18 at 17:06
J.G.J.G.
25.6k22539
$endgroup$
add a comment |
$begingroup$
One long line proves it: $$int_a^b |u(t)|^2 dt=int_a^bleft|int_a^t 1cdot u'(t')dt'right|^2 dtleint_a^bleft[int_a^t 1^2 dt'cdotint_a^t |u'(t')|^2dt'right] dt\leint_a^bleft[(t-a)cdotint_a^b |u'(t')|^2dt'right] dt=frac{(b-a)^2}{2}int_a^b |u'(t')|^2dt'.$$The first $le$ uses Cauchy-Schwarz; the second replaces an $int_a^t dt'$ with $int_a^b dt'$.
answered Dec 22 '18 at 17:06
J.G.J.G.
25.6k22539
$endgroup$
One long line proves it: $$int_a^b |u(t)|^2 dt=int_a^bleft|int_a^t 1cdot u'(t')dt'right|^2 dtleint_a^bleft[int_a^t 1^2 dt'cdotint_a^t |u'(t')|^2dt'right] dt\leint_a^bleft[(t-a)cdotint_a^b |u'(t')|^2dt'right] dt=frac{(b-a)^2}{2}int_a^b |u'(t')|^2dt'.$$The first $le$ uses Cauchy-Schwarz; the second replaces an $int_a^t dt'$ with $int_a^b dt'$.
answered Dec 22 '18 at 17:06
J.G.J.G.
25.6k22539
answered Dec 22 '18 at 17:06
J.G.J.G.
25.6k22539
answered Dec 22 '18 at 17:06
J.G.J.G.
25.6k22539
answered Dec 22 '18 at 17:06
J.G.J.G.
25.6k22539
25.6k22539
add a comment |
add a comment |
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 16:47
$begingroup$
There is no conclusion after ‘then’, only a formula.
$endgroup$
– Bernard
Dec 22 '18 at 16:52