Integral of the product of two functions equals zero
$begingroup$
Let $alpha:[a,b]to mathbb{R}$ a continuous function such that
$$ int_{a}^{b}alpha(x)h(x) ; dx=0$$ for any continuous function $h:[a,b]to mathbb{R}$ such that $h(a)=h(b)=0$. How can I prove that $alpha(x)=0$ for all $xin [a,b]$. Any help will be very appreciated.
real-analysis calculus
asked Dec 19 '18 at 23:39
Diego VargasDiego Vargas
1227
$endgroup$
add a comment |
$begingroup$
Let $alpha:[a,b]to mathbb{R}$ a continuous function such that
$$ int_{a}^{b}alpha(x)h(x) ; dx=0$$ for any continuous function $h:[a,b]to mathbb{R}$ such that $h(a)=h(b)=0$. How can I prove that $alpha(x)=0$ for all $xin [a,b]$. Any help will be very appreciated.
real-analysis calculus
asked Dec 19 '18 at 23:39
Diego VargasDiego Vargas
1227
$endgroup$
add a comment |
$begingroup$
Let $alpha:[a,b]to mathbb{R}$ a continuous function such that
$$ int_{a}^{b}alpha(x)h(x) ; dx=0$$ for any continuous function $h:[a,b]to mathbb{R}$ such that $h(a)=h(b)=0$. How can I prove that $alpha(x)=0$ for all $xin [a,b]$. Any help will be very appreciated.
real-analysis calculus
asked Dec 19 '18 at 23:39
Diego VargasDiego Vargas
1227
$endgroup$
Let $alpha:[a,b]to mathbb{R}$ a continuous function such that
$$ int_{a}^{b}alpha(x)h(x) ; dx=0$$ for any continuous function $h:[a,b]to mathbb{R}$ such that $h(a)=h(b)=0$. How can I prove that $alpha(x)=0$ for all $xin [a,b]$. Any help will be very appreciated.
real-analysis calculus
real-analysis calculus
asked Dec 19 '18 at 23:39
Diego VargasDiego Vargas
1227
asked Dec 19 '18 at 23:39
Diego VargasDiego Vargas
1227
asked Dec 19 '18 at 23:39
Diego VargasDiego Vargas
1227
asked Dec 19 '18 at 23:39
Diego VargasDiego Vargas
1227
asked Dec 19 '18 at 23:39
Diego VargasDiego Vargas
1227
1227
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
HINT: Suppose $alpha(x_0)ne 0$, and so $alpha$ is nonzero (say positive) on some interval $(x_0-delta,x_0+delta)$. Can you make up an appropriate function $h$ that will give you a nonzero integral?
answered Dec 19 '18 at 23:42
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
$begingroup$
I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
$endgroup$
– Diego Vargas
Dec 19 '18 at 23:46
1
$begingroup$
Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
$endgroup$
– John Hughes
Dec 19 '18 at 23:48
add a comment |
$begingroup$
Let $h(x)=alpha (x)$ for $a+epsilon <x <b-epsilon$, $h(a)=h(b)=0$ and let $h$ have straight line graphs in $[a,a+epsilon]$ as well as $[b-epsilon,b]$. Write down the given equation and let $epsilon to 0$. You will get $int (alpha (x))^{2}), dx=0$ so $alpha =0$. [ Note that $h$ is bounded by $sup {|alpha (x)|: aleq x leq b}=M$ say. Hence $int _a ^{a+epsilon} |alpha (x) h(x)| , dxleq M^{2} epsilon to 0$. Similar argument for integral over $[b-epsilon,b]$].
answered Dec 19 '18 at 23:50
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
add a comment |
$begingroup$
Consider proving this by contradiction, where you assume that $alpha (x_0) neq 0$ for some $x_0 in [a,b]$, then due to $alpha$ being continuous, this means there is some region (often called a "ball") around $x_0$ where $alpha$ is non-zero, even if $x_0$ is at a boundary, i.e., $a$ or $b$. Next, think about how that will affect the integration using any continuous function $h(x)$, especially if its values are very large or small in that region, especially compared to its values elsewhere in the range $[a,b]$.
answered Dec 19 '18 at 23:48
John OmielanJohn Omielan
4,6312215
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046984%2fintegral-of-the-product-of-two-functions-equals-zero%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT: Suppose $alpha(x_0)ne 0$, and so $alpha$ is nonzero (say positive) on some interval $(x_0-delta,x_0+delta)$. Can you make up an appropriate function $h$ that will give you a nonzero integral?
answered Dec 19 '18 at 23:42
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
$begingroup$
I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
$endgroup$
– Diego Vargas
Dec 19 '18 at 23:46
1
$begingroup$
Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
$endgroup$
– John Hughes
Dec 19 '18 at 23:48
add a comment |
$begingroup$
HINT: Suppose $alpha(x_0)ne 0$, and so $alpha$ is nonzero (say positive) on some interval $(x_0-delta,x_0+delta)$. Can you make up an appropriate function $h$ that will give you a nonzero integral?
answered Dec 19 '18 at 23:42
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
$begingroup$
I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
$endgroup$
– Diego Vargas
Dec 19 '18 at 23:46
1
$begingroup$
Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
$endgroup$
– John Hughes
Dec 19 '18 at 23:48
add a comment |
$begingroup$
HINT: Suppose $alpha(x_0)ne 0$, and so $alpha$ is nonzero (say positive) on some interval $(x_0-delta,x_0+delta)$. Can you make up an appropriate function $h$ that will give you a nonzero integral?
answered Dec 19 '18 at 23:42
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
HINT: Suppose $alpha(x_0)ne 0$, and so $alpha$ is nonzero (say positive) on some interval $(x_0-delta,x_0+delta)$. Can you make up an appropriate function $h$ that will give you a nonzero integral?
answered Dec 19 '18 at 23:42
Ted ShifrinTed Shifrin
64.7k44692
answered Dec 19 '18 at 23:42
Ted ShifrinTed Shifrin
64.7k44692
answered Dec 19 '18 at 23:42
Ted ShifrinTed Shifrin
64.7k44692
answered Dec 19 '18 at 23:42
Ted ShifrinTed Shifrin
64.7k44692
64.7k44692
$begingroup$
I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
$endgroup$
– Diego Vargas
Dec 19 '18 at 23:46
1
$begingroup$
Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
$endgroup$
– John Hughes
Dec 19 '18 at 23:48
add a comment |
$begingroup$
I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
$endgroup$
– Diego Vargas
Dec 19 '18 at 23:46
1
$begingroup$
Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
$endgroup$
– John Hughes
Dec 19 '18 at 23:48
$begingroup$
I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
$endgroup$
– Diego Vargas
Dec 19 '18 at 23:46
$begingroup$
I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
$endgroup$
– Diego Vargas
Dec 19 '18 at 23:46
1
1
$begingroup$
Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
$endgroup$
– John Hughes
Dec 19 '18 at 23:48
$begingroup$
Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
$endgroup$
– John Hughes
Dec 19 '18 at 23:48
add a comment |
$begingroup$
Let $h(x)=alpha (x)$ for $a+epsilon <x <b-epsilon$, $h(a)=h(b)=0$ and let $h$ have straight line graphs in $[a,a+epsilon]$ as well as $[b-epsilon,b]$. Write down the given equation and let $epsilon to 0$. You will get $int (alpha (x))^{2}), dx=0$ so $alpha =0$. [ Note that $h$ is bounded by $sup {|alpha (x)|: aleq x leq b}=M$ say. Hence $int _a ^{a+epsilon} |alpha (x) h(x)| , dxleq M^{2} epsilon to 0$. Similar argument for integral over $[b-epsilon,b]$].
answered Dec 19 '18 at 23:50
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
add a comment |
$begingroup$
Let $h(x)=alpha (x)$ for $a+epsilon <x <b-epsilon$, $h(a)=h(b)=0$ and let $h$ have straight line graphs in $[a,a+epsilon]$ as well as $[b-epsilon,b]$. Write down the given equation and let $epsilon to 0$. You will get $int (alpha (x))^{2}), dx=0$ so $alpha =0$. [ Note that $h$ is bounded by $sup {|alpha (x)|: aleq x leq b}=M$ say. Hence $int _a ^{a+epsilon} |alpha (x) h(x)| , dxleq M^{2} epsilon to 0$. Similar argument for integral over $[b-epsilon,b]$].
answered Dec 19 '18 at 23:50
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
add a comment |
$begingroup$
Let $h(x)=alpha (x)$ for $a+epsilon <x <b-epsilon$, $h(a)=h(b)=0$ and let $h$ have straight line graphs in $[a,a+epsilon]$ as well as $[b-epsilon,b]$. Write down the given equation and let $epsilon to 0$. You will get $int (alpha (x))^{2}), dx=0$ so $alpha =0$. [ Note that $h$ is bounded by $sup {|alpha (x)|: aleq x leq b}=M$ say. Hence $int _a ^{a+epsilon} |alpha (x) h(x)| , dxleq M^{2} epsilon to 0$. Similar argument for integral over $[b-epsilon,b]$].
answered Dec 19 '18 at 23:50
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
Let $h(x)=alpha (x)$ for $a+epsilon <x <b-epsilon$, $h(a)=h(b)=0$ and let $h$ have straight line graphs in $[a,a+epsilon]$ as well as $[b-epsilon,b]$. Write down the given equation and let $epsilon to 0$. You will get $int (alpha (x))^{2}), dx=0$ so $alpha =0$. [ Note that $h$ is bounded by $sup {|alpha (x)|: aleq x leq b}=M$ say. Hence $int _a ^{a+epsilon} |alpha (x) h(x)| , dxleq M^{2} epsilon to 0$. Similar argument for integral over $[b-epsilon,b]$].
answered Dec 19 '18 at 23:50
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Dec 19 '18 at 23:50
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Dec 19 '18 at 23:50
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Dec 19 '18 at 23:50
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
72.6k53170
add a comment |
add a comment |
$begingroup$
Consider proving this by contradiction, where you assume that $alpha (x_0) neq 0$ for some $x_0 in [a,b]$, then due to $alpha$ being continuous, this means there is some region (often called a "ball") around $x_0$ where $alpha$ is non-zero, even if $x_0$ is at a boundary, i.e., $a$ or $b$. Next, think about how that will affect the integration using any continuous function $h(x)$, especially if its values are very large or small in that region, especially compared to its values elsewhere in the range $[a,b]$.
answered Dec 19 '18 at 23:48
John OmielanJohn Omielan
4,6312215
$endgroup$
add a comment |
$begingroup$
Consider proving this by contradiction, where you assume that $alpha (x_0) neq 0$ for some $x_0 in [a,b]$, then due to $alpha$ being continuous, this means there is some region (often called a "ball") around $x_0$ where $alpha$ is non-zero, even if $x_0$ is at a boundary, i.e., $a$ or $b$. Next, think about how that will affect the integration using any continuous function $h(x)$, especially if its values are very large or small in that region, especially compared to its values elsewhere in the range $[a,b]$.
answered Dec 19 '18 at 23:48
John OmielanJohn Omielan
4,6312215
$endgroup$
add a comment |
$begingroup$
Consider proving this by contradiction, where you assume that $alpha (x_0) neq 0$ for some $x_0 in [a,b]$, then due to $alpha$ being continuous, this means there is some region (often called a "ball") around $x_0$ where $alpha$ is non-zero, even if $x_0$ is at a boundary, i.e., $a$ or $b$. Next, think about how that will affect the integration using any continuous function $h(x)$, especially if its values are very large or small in that region, especially compared to its values elsewhere in the range $[a,b]$.
answered Dec 19 '18 at 23:48
John OmielanJohn Omielan
4,6312215
$endgroup$
Consider proving this by contradiction, where you assume that $alpha (x_0) neq 0$ for some $x_0 in [a,b]$, then due to $alpha$ being continuous, this means there is some region (often called a "ball") around $x_0$ where $alpha$ is non-zero, even if $x_0$ is at a boundary, i.e., $a$ or $b$. Next, think about how that will affect the integration using any continuous function $h(x)$, especially if its values are very large or small in that region, especially compared to its values elsewhere in the range $[a,b]$.
answered Dec 19 '18 at 23:48
John OmielanJohn Omielan
4,6312215
answered Dec 19 '18 at 23:48
John OmielanJohn Omielan
4,6312215
answered Dec 19 '18 at 23:48
John OmielanJohn Omielan
4,6312215
answered Dec 19 '18 at 23:48
John OmielanJohn Omielan
4,6312215
4,6312215
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046984%2fintegral-of-the-product-of-two-functions-equals-zero%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
