“Quick” proof of the fundamental lemma of calculus of variationsWhy can't we construct a counter-example to the Fundamental Lemma of the Calculus of Variations?Fundamental lemma of calculus of variation .A variation of fundamental lemma of variation of calculus .Variation of the fundamental lemma of calculus of variationProof of fundamental lemma of calculus of variation.Proof of Fundamental Lemma of Calculus of VariationsFundamental lemma of calculus of variation, about hypothesisWeakening the Fundamental Lemma of Calculus of VariationsFundamental Lemma for variational calculus for measurable functionCompact support in calculus of variations
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“Quick” proof of the fundamental lemma of calculus of variations
Why can't we construct a counter-example to the Fundamental Lemma of the Calculus of Variations?Fundamental lemma of calculus of variation .A variation of fundamental lemma of variation of calculus .Variation of the fundamental lemma of calculus of variationProof of fundamental lemma of calculus of variation.Proof of Fundamental Lemma of Calculus of VariationsFundamental lemma of calculus of variation, about hypothesisWeakening the Fundamental Lemma of Calculus of VariationsFundamental Lemma for variational calculus for measurable functionCompact support in calculus of variations
$begingroup$
Here's the statement:
Let $f in C([a,b])$ and $H$ be the set $hin C([a,b]):h(a)=h(b)=0$. If $int_a^bf(x)h(x),textdx=0$ for all $hin H$, then $f(x)=0$ for all $xin [a,b]$.
I saw a lot of long proofs for this but I thought I could do better, but I think there could be a subtle error. Here's my attempt:
Consider the constant $c$ and the function $phi$ defined as
$$phi(x)=int_a^xh(x)-c,textdx, c=frac1b-aint_a^bh(x),textdx$$
obviously $phi(a)=phi(b)=0$, hence $phi in H$. Also, $h$ is integrable because it is continuous in $[a,b]$.
By hypothesis, we have
$$int_a^bf(x)h(x),textdx=0Leftrightarrowint_a^bf(x)phi'(x),textdx=0$$
Then we can use Du Bois-Reymond's lemma, which states
Let $H$ be the set $hin C^1([a,b]):h(a)=h(b)=0$. If $fin C([a,b])$ and $int_a^b f(x)h'(x),textdx=0$ for all $hin H$, then $f(x)$ is constant for all $xin[a,b]$.
The lemma can be used directly to get that $f(x)=k$, where $kinmathbbR$.
Then our hypothesis is simply
$$int_a^bkh(x),textdx=0$$
for this to be true for all $h in H$, we must have $k=0$, because if $h(x)$ were, for example, a positive function in $(a,b)$ with $h(a)=h(b)=0$, it would not be true... thus the lemma is proved.
Now, where does it all fall apart?
proof-verification calculus-of-variations
asked Mar 13 at 18:26
AstlyDichrarAstlyDichrar
42248
$endgroup$
add a comment |
$begingroup$
Here's the statement:
Let $f in C([a,b])$ and $H$ be the set $hin C([a,b]):h(a)=h(b)=0$. If $int_a^bf(x)h(x),textdx=0$ for all $hin H$, then $f(x)=0$ for all $xin [a,b]$.
I saw a lot of long proofs for this but I thought I could do better, but I think there could be a subtle error. Here's my attempt:
Consider the constant $c$ and the function $phi$ defined as
$$phi(x)=int_a^xh(x)-c,textdx, c=frac1b-aint_a^bh(x),textdx$$
obviously $phi(a)=phi(b)=0$, hence $phi in H$. Also, $h$ is integrable because it is continuous in $[a,b]$.
By hypothesis, we have
$$int_a^bf(x)h(x),textdx=0Leftrightarrowint_a^bf(x)phi'(x),textdx=0$$
Then we can use Du Bois-Reymond's lemma, which states
Let $H$ be the set $hin C^1([a,b]):h(a)=h(b)=0$. If $fin C([a,b])$ and $int_a^b f(x)h'(x),textdx=0$ for all $hin H$, then $f(x)$ is constant for all $xin[a,b]$.
The lemma can be used directly to get that $f(x)=k$, where $kinmathbbR$.
Then our hypothesis is simply
$$int_a^bkh(x),textdx=0$$
for this to be true for all $h in H$, we must have $k=0$, because if $h(x)$ were, for example, a positive function in $(a,b)$ with $h(a)=h(b)=0$, it would not be true... thus the lemma is proved.
Now, where does it all fall apart?
proof-verification calculus-of-variations
asked Mar 13 at 18:26
AstlyDichrarAstlyDichrar
42248
$endgroup$
$begingroup$
Don't you need the first statement to prove the lemma? It looks like the lemma follows from the first using integration by parts (do you know any other way to deduce it? because otherwise it is a circular proof).
$endgroup$
– Yanko
Mar 13 at 18:31
$begingroup$
The proof is very similar. You cannot use integration by parts if the function $f$ in Du Bois-Raymond's lemma is not in $C^1([a,b])$, but it is provable without it. Essentially, I'm construction the function $phi$ because it is in $C^1([a,b])$, where I can apply Du-Bois Raymond ($h$ is not $C^1([a,b])$ in the fundamental lemma).
$endgroup$
– AstlyDichrar
Mar 13 at 18:38
$begingroup$
To add on to this, there is a proof for Du Bois-Reymond that does not depend on any of what I did.
$endgroup$
– AstlyDichrar
Mar 13 at 18:45
$begingroup$
I see, well in that case I don't see any flaws with your proof.
$endgroup$
– Yanko
Mar 13 at 18:46
add a comment |
$begingroup$
Here's the statement:
Let $f in C([a,b])$ and $H$ be the set $hin C([a,b]):h(a)=h(b)=0$. If $int_a^bf(x)h(x),textdx=0$ for all $hin H$, then $f(x)=0$ for all $xin [a,b]$.
I saw a lot of long proofs for this but I thought I could do better, but I think there could be a subtle error. Here's my attempt:
Consider the constant $c$ and the function $phi$ defined as
$$phi(x)=int_a^xh(x)-c,textdx, c=frac1b-aint_a^bh(x),textdx$$
obviously $phi(a)=phi(b)=0$, hence $phi in H$. Also, $h$ is integrable because it is continuous in $[a,b]$.
By hypothesis, we have
$$int_a^bf(x)h(x),textdx=0Leftrightarrowint_a^bf(x)phi'(x),textdx=0$$
Then we can use Du Bois-Reymond's lemma, which states
Let $H$ be the set $hin C^1([a,b]):h(a)=h(b)=0$. If $fin C([a,b])$ and $int_a^b f(x)h'(x),textdx=0$ for all $hin H$, then $f(x)$ is constant for all $xin[a,b]$.
The lemma can be used directly to get that $f(x)=k$, where $kinmathbbR$.
Then our hypothesis is simply
$$int_a^bkh(x),textdx=0$$
for this to be true for all $h in H$, we must have $k=0$, because if $h(x)$ were, for example, a positive function in $(a,b)$ with $h(a)=h(b)=0$, it would not be true... thus the lemma is proved.
Now, where does it all fall apart?
proof-verification calculus-of-variations
asked Mar 13 at 18:26
AstlyDichrarAstlyDichrar
42248
$endgroup$
Here's the statement:
Let $f in C([a,b])$ and $H$ be the set $hin C([a,b]):h(a)=h(b)=0$. If $int_a^bf(x)h(x),textdx=0$ for all $hin H$, then $f(x)=0$ for all $xin [a,b]$.
I saw a lot of long proofs for this but I thought I could do better, but I think there could be a subtle error. Here's my attempt:
Consider the constant $c$ and the function $phi$ defined as
$$phi(x)=int_a^xh(x)-c,textdx, c=frac1b-aint_a^bh(x),textdx$$
obviously $phi(a)=phi(b)=0$, hence $phi in H$. Also, $h$ is integrable because it is continuous in $[a,b]$.
By hypothesis, we have
$$int_a^bf(x)h(x),textdx=0Leftrightarrowint_a^bf(x)phi'(x),textdx=0$$
Then we can use Du Bois-Reymond's lemma, which states
Let $H$ be the set $hin C^1([a,b]):h(a)=h(b)=0$. If $fin C([a,b])$ and $int_a^b f(x)h'(x),textdx=0$ for all $hin H$, then $f(x)$ is constant for all $xin[a,b]$.
The lemma can be used directly to get that $f(x)=k$, where $kinmathbbR$.
Then our hypothesis is simply
$$int_a^bkh(x),textdx=0$$
for this to be true for all $h in H$, we must have $k=0$, because if $h(x)$ were, for example, a positive function in $(a,b)$ with $h(a)=h(b)=0$, it would not be true... thus the lemma is proved.
Now, where does it all fall apart?
proof-verification calculus-of-variations
proof-verification calculus-of-variations
asked Mar 13 at 18:26
AstlyDichrarAstlyDichrar
42248
asked Mar 13 at 18:26
AstlyDichrarAstlyDichrar
42248
asked Mar 13 at 18:26
AstlyDichrarAstlyDichrar
42248
asked Mar 13 at 18:26
AstlyDichrarAstlyDichrar
42248
asked Mar 13 at 18:26
AstlyDichrarAstlyDichrar
42248
42248
$begingroup$
Don't you need the first statement to prove the lemma? It looks like the lemma follows from the first using integration by parts (do you know any other way to deduce it? because otherwise it is a circular proof).
$endgroup$
– Yanko
Mar 13 at 18:31
$begingroup$
The proof is very similar. You cannot use integration by parts if the function $f$ in Du Bois-Raymond's lemma is not in $C^1([a,b])$, but it is provable without it. Essentially, I'm construction the function $phi$ because it is in $C^1([a,b])$, where I can apply Du-Bois Raymond ($h$ is not $C^1([a,b])$ in the fundamental lemma).
$endgroup$
– AstlyDichrar
Mar 13 at 18:38
$begingroup$
To add on to this, there is a proof for Du Bois-Reymond that does not depend on any of what I did.
$endgroup$
– AstlyDichrar
Mar 13 at 18:45
$begingroup$
I see, well in that case I don't see any flaws with your proof.
$endgroup$
– Yanko
Mar 13 at 18:46
add a comment |
$begingroup$
Don't you need the first statement to prove the lemma? It looks like the lemma follows from the first using integration by parts (do you know any other way to deduce it? because otherwise it is a circular proof).
$endgroup$
– Yanko
Mar 13 at 18:31
$begingroup$
The proof is very similar. You cannot use integration by parts if the function $f$ in Du Bois-Raymond's lemma is not in $C^1([a,b])$, but it is provable without it. Essentially, I'm construction the function $phi$ because it is in $C^1([a,b])$, where I can apply Du-Bois Raymond ($h$ is not $C^1([a,b])$ in the fundamental lemma).
$endgroup$
– AstlyDichrar
Mar 13 at 18:38
$begingroup$
To add on to this, there is a proof for Du Bois-Reymond that does not depend on any of what I did.
$endgroup$
– AstlyDichrar
Mar 13 at 18:45
$begingroup$
I see, well in that case I don't see any flaws with your proof.
$endgroup$
– Yanko
Mar 13 at 18:46
$begingroup$
Don't you need the first statement to prove the lemma? It looks like the lemma follows from the first using integration by parts (do you know any other way to deduce it? because otherwise it is a circular proof).
$endgroup$
– Yanko
Mar 13 at 18:31
$begingroup$
Don't you need the first statement to prove the lemma? It looks like the lemma follows from the first using integration by parts (do you know any other way to deduce it? because otherwise it is a circular proof).
$endgroup$
– Yanko
Mar 13 at 18:31
$begingroup$
The proof is very similar. You cannot use integration by parts if the function $f$ in Du Bois-Raymond's lemma is not in $C^1([a,b])$, but it is provable without it. Essentially, I'm construction the function $phi$ because it is in $C^1([a,b])$, where I can apply Du-Bois Raymond ($h$ is not $C^1([a,b])$ in the fundamental lemma).
$endgroup$
– AstlyDichrar
Mar 13 at 18:38
$begingroup$
The proof is very similar. You cannot use integration by parts if the function $f$ in Du Bois-Raymond's lemma is not in $C^1([a,b])$, but it is provable without it. Essentially, I'm construction the function $phi$ because it is in $C^1([a,b])$, where I can apply Du-Bois Raymond ($h$ is not $C^1([a,b])$ in the fundamental lemma).
$endgroup$
– AstlyDichrar
Mar 13 at 18:38
$begingroup$
To add on to this, there is a proof for Du Bois-Reymond that does not depend on any of what I did.
$endgroup$
– AstlyDichrar
Mar 13 at 18:45
$begingroup$
To add on to this, there is a proof for Du Bois-Reymond that does not depend on any of what I did.
$endgroup$
– AstlyDichrar
Mar 13 at 18:45
$begingroup$
I see, well in that case I don't see any flaws with your proof.
$endgroup$
– Yanko
Mar 13 at 18:46
$begingroup$
I see, well in that case I don't see any flaws with your proof.
$endgroup$
– Yanko
Mar 13 at 18:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Alright, my professor pointed out a possible flaw in the proof: is it really possible to represent every function in $H = h in C^1([a,b]):h(a)=h(b)=0$ as $phi(x)$, given its definition in the original post?
I'd guess that means we need to verify whether or not for every function $hin H$ ($H$ in the sense of the original definition, where we consider functions in $C([a,b])$, there is a correspondent function $fin H$ (in the $C^1([a,b])$ sense) such that $f(x)=int_a^b h(x)-c,textdx$. Somewhat informally, we need to verify if this integral is a "surjection", I think. This is probably not true, but I don't think I know enough to prove it.
answered 7 hours ago
AstlyDichrarAstlyDichrar
42248
$endgroup$
add a comment |
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$begingroup$
Alright, my professor pointed out a possible flaw in the proof: is it really possible to represent every function in $H = h in C^1([a,b]):h(a)=h(b)=0$ as $phi(x)$, given its definition in the original post?
I'd guess that means we need to verify whether or not for every function $hin H$ ($H$ in the sense of the original definition, where we consider functions in $C([a,b])$, there is a correspondent function $fin H$ (in the $C^1([a,b])$ sense) such that $f(x)=int_a^b h(x)-c,textdx$. Somewhat informally, we need to verify if this integral is a "surjection", I think. This is probably not true, but I don't think I know enough to prove it.
answered 7 hours ago
AstlyDichrarAstlyDichrar
42248
$endgroup$
add a comment |
$begingroup$
Alright, my professor pointed out a possible flaw in the proof: is it really possible to represent every function in $H = h in C^1([a,b]):h(a)=h(b)=0$ as $phi(x)$, given its definition in the original post?
I'd guess that means we need to verify whether or not for every function $hin H$ ($H$ in the sense of the original definition, where we consider functions in $C([a,b])$, there is a correspondent function $fin H$ (in the $C^1([a,b])$ sense) such that $f(x)=int_a^b h(x)-c,textdx$. Somewhat informally, we need to verify if this integral is a "surjection", I think. This is probably not true, but I don't think I know enough to prove it.
answered 7 hours ago
AstlyDichrarAstlyDichrar
42248
$endgroup$
add a comment |
$begingroup$
Alright, my professor pointed out a possible flaw in the proof: is it really possible to represent every function in $H = h in C^1([a,b]):h(a)=h(b)=0$ as $phi(x)$, given its definition in the original post?
I'd guess that means we need to verify whether or not for every function $hin H$ ($H$ in the sense of the original definition, where we consider functions in $C([a,b])$, there is a correspondent function $fin H$ (in the $C^1([a,b])$ sense) such that $f(x)=int_a^b h(x)-c,textdx$. Somewhat informally, we need to verify if this integral is a "surjection", I think. This is probably not true, but I don't think I know enough to prove it.
answered 7 hours ago
AstlyDichrarAstlyDichrar
42248
$endgroup$
Alright, my professor pointed out a possible flaw in the proof: is it really possible to represent every function in $H = h in C^1([a,b]):h(a)=h(b)=0$ as $phi(x)$, given its definition in the original post?
I'd guess that means we need to verify whether or not for every function $hin H$ ($H$ in the sense of the original definition, where we consider functions in $C([a,b])$, there is a correspondent function $fin H$ (in the $C^1([a,b])$ sense) such that $f(x)=int_a^b h(x)-c,textdx$. Somewhat informally, we need to verify if this integral is a "surjection", I think. This is probably not true, but I don't think I know enough to prove it.
answered 7 hours ago
AstlyDichrarAstlyDichrar
42248
answered 7 hours ago
AstlyDichrarAstlyDichrar
42248
answered 7 hours ago
AstlyDichrarAstlyDichrar
42248
answered 7 hours ago
AstlyDichrarAstlyDichrar
42248
42248
add a comment |
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$begingroup$
Don't you need the first statement to prove the lemma? It looks like the lemma follows from the first using integration by parts (do you know any other way to deduce it? because otherwise it is a circular proof).
$endgroup$
– Yanko
Mar 13 at 18:31
$begingroup$
The proof is very similar. You cannot use integration by parts if the function $f$ in Du Bois-Raymond's lemma is not in $C^1([a,b])$, but it is provable without it. Essentially, I'm construction the function $phi$ because it is in $C^1([a,b])$, where I can apply Du-Bois Raymond ($h$ is not $C^1([a,b])$ in the fundamental lemma).
$endgroup$
– AstlyDichrar
Mar 13 at 18:38
$begingroup$
To add on to this, there is a proof for Du Bois-Reymond that does not depend on any of what I did.
$endgroup$
– AstlyDichrar
Mar 13 at 18:45
$begingroup$
I see, well in that case I don't see any flaws with your proof.
$endgroup$
– Yanko
Mar 13 at 18:46