Part of proof that supremum norm is (positive) definite. $||f||_infty =0 implies f=0$.
$begingroup$
I want to prove that for a bounded function $f$, we have:
$$ ||f||_infty =0 implies f=0. $$
Simply observe that if we denote the domain of definition by $D$:
$$ ||f||_infty = sup_{x in D}|f(x)|$$
We know that the supremum is the least upper bound, so for any $y in D $:
$$ |f(y)| leq sup_{x in D} |f(x)|=0$$
But we also know that the Euclidean metric is positive definite, so
$|f(y) |geq 0$, and we get that for all $y in D$ we have $|f(y)|=0$, but now again by positive definiteness:
$$|f(y)|=0 iff f(y)=0.$$
Is my reasoning correct?
real-analysis proof-verification supremum-and-infimum
asked Dec 16 '18 at 21:30
Wesley StrikWesley Strik
2,017423
$endgroup$
add a comment |
$begingroup$
I want to prove that for a bounded function $f$, we have:
$$ ||f||_infty =0 implies f=0. $$
Simply observe that if we denote the domain of definition by $D$:
$$ ||f||_infty = sup_{x in D}|f(x)|$$
We know that the supremum is the least upper bound, so for any $y in D $:
$$ |f(y)| leq sup_{x in D} |f(x)|=0$$
But we also know that the Euclidean metric is positive definite, so
$|f(y) |geq 0$, and we get that for all $y in D$ we have $|f(y)|=0$, but now again by positive definiteness:
$$|f(y)|=0 iff f(y)=0.$$
Is my reasoning correct?
real-analysis proof-verification supremum-and-infimum
asked Dec 16 '18 at 21:30
Wesley StrikWesley Strik
2,017423
$endgroup$
1
$begingroup$
It's correct, but for a proof this simple I would want to see the dummy variable that is being dropped from the argument: If $Vert fVert_{infty}=0$ then $0=Vert fVert_{infty}=sup_{x}|f(x)|geq|f(y)|$ for any $y$.
$endgroup$
– parsiad
Dec 16 '18 at 21:38
1
$begingroup$
This is Quite OK.
$endgroup$
– szw1710
Dec 16 '18 at 21:38
$begingroup$
Is this what you meant? I think that's a useful addition parsiad.
$endgroup$
– Wesley Strik
Dec 16 '18 at 21:40
add a comment |
$begingroup$
I want to prove that for a bounded function $f$, we have:
$$ ||f||_infty =0 implies f=0. $$
Simply observe that if we denote the domain of definition by $D$:
$$ ||f||_infty = sup_{x in D}|f(x)|$$
We know that the supremum is the least upper bound, so for any $y in D $:
$$ |f(y)| leq sup_{x in D} |f(x)|=0$$
But we also know that the Euclidean metric is positive definite, so
$|f(y) |geq 0$, and we get that for all $y in D$ we have $|f(y)|=0$, but now again by positive definiteness:
$$|f(y)|=0 iff f(y)=0.$$
Is my reasoning correct?
real-analysis proof-verification supremum-and-infimum
asked Dec 16 '18 at 21:30
Wesley StrikWesley Strik
2,017423
$endgroup$
I want to prove that for a bounded function $f$, we have:
$$ ||f||_infty =0 implies f=0. $$
Simply observe that if we denote the domain of definition by $D$:
$$ ||f||_infty = sup_{x in D}|f(x)|$$
We know that the supremum is the least upper bound, so for any $y in D $:
$$ |f(y)| leq sup_{x in D} |f(x)|=0$$
But we also know that the Euclidean metric is positive definite, so
$|f(y) |geq 0$, and we get that for all $y in D$ we have $|f(y)|=0$, but now again by positive definiteness:
$$|f(y)|=0 iff f(y)=0.$$
Is my reasoning correct?
real-analysis proof-verification supremum-and-infimum
real-analysis proof-verification supremum-and-infimum
asked Dec 16 '18 at 21:30
Wesley StrikWesley Strik
2,017423
asked Dec 16 '18 at 21:30
Wesley StrikWesley Strik
2,017423
edited Dec 16 '18 at 22:35
Wesley Strik
asked Dec 16 '18 at 21:30
Wesley StrikWesley Strik
2,017423
asked Dec 16 '18 at 21:30
Wesley StrikWesley Strik
2,017423
asked Dec 16 '18 at 21:30
Wesley StrikWesley Strik
2,017423
2,017423
1
$begingroup$
It's correct, but for a proof this simple I would want to see the dummy variable that is being dropped from the argument: If $Vert fVert_{infty}=0$ then $0=Vert fVert_{infty}=sup_{x}|f(x)|geq|f(y)|$ for any $y$.
$endgroup$
– parsiad
Dec 16 '18 at 21:38
1
$begingroup$
This is Quite OK.
$endgroup$
– szw1710
Dec 16 '18 at 21:38
$begingroup$
Is this what you meant? I think that's a useful addition parsiad.
$endgroup$
– Wesley Strik
Dec 16 '18 at 21:40
add a comment |
1
$begingroup$
It's correct, but for a proof this simple I would want to see the dummy variable that is being dropped from the argument: If $Vert fVert_{infty}=0$ then $0=Vert fVert_{infty}=sup_{x}|f(x)|geq|f(y)|$ for any $y$.
$endgroup$
– parsiad
Dec 16 '18 at 21:38
1
$begingroup$
This is Quite OK.
$endgroup$
– szw1710
Dec 16 '18 at 21:38
$begingroup$
Is this what you meant? I think that's a useful addition parsiad.
$endgroup$
– Wesley Strik
Dec 16 '18 at 21:40
1
1
$begingroup$
It's correct, but for a proof this simple I would want to see the dummy variable that is being dropped from the argument: If $Vert fVert_{infty}=0$ then $0=Vert fVert_{infty}=sup_{x}|f(x)|geq|f(y)|$ for any $y$.
$endgroup$
– parsiad
Dec 16 '18 at 21:38
$begingroup$
It's correct, but for a proof this simple I would want to see the dummy variable that is being dropped from the argument: If $Vert fVert_{infty}=0$ then $0=Vert fVert_{infty}=sup_{x}|f(x)|geq|f(y)|$ for any $y$.
$endgroup$
– parsiad
Dec 16 '18 at 21:38
1
1
$begingroup$
This is Quite OK.
$endgroup$
– szw1710
Dec 16 '18 at 21:38
$begingroup$
This is Quite OK.
$endgroup$
– szw1710
Dec 16 '18 at 21:38
$begingroup$
Is this what you meant? I think that's a useful addition parsiad.
$endgroup$
– Wesley Strik
Dec 16 '18 at 21:40
$begingroup$
Is this what you meant? I think that's a useful addition parsiad.
$endgroup$
– Wesley Strik
Dec 16 '18 at 21:40
add a comment |
1 Answer
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$begingroup$
Yes it is correct, alternatively, as suggested by Quiliup, We can immediately conclude that since for all $y in D$ $$|f(y)|leq 0$$
By positive definiteness of the Euclidean norm it must hold that for all $y in D$ we have $|f(y)|=0$ and hence $f(y)=0$.
answered Dec 16 '18 at 23:16
Wesley StrikWesley Strik
2,017423
$endgroup$
add a comment |
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$begingroup$
Yes it is correct, alternatively, as suggested by Quiliup, We can immediately conclude that since for all $y in D$ $$|f(y)|leq 0$$
By positive definiteness of the Euclidean norm it must hold that for all $y in D$ we have $|f(y)|=0$ and hence $f(y)=0$.
answered Dec 16 '18 at 23:16
Wesley StrikWesley Strik
2,017423
$endgroup$
add a comment |
$begingroup$
Yes it is correct, alternatively, as suggested by Quiliup, We can immediately conclude that since for all $y in D$ $$|f(y)|leq 0$$
By positive definiteness of the Euclidean norm it must hold that for all $y in D$ we have $|f(y)|=0$ and hence $f(y)=0$.
answered Dec 16 '18 at 23:16
Wesley StrikWesley Strik
2,017423
$endgroup$
add a comment |
$begingroup$
Yes it is correct, alternatively, as suggested by Quiliup, We can immediately conclude that since for all $y in D$ $$|f(y)|leq 0$$
By positive definiteness of the Euclidean norm it must hold that for all $y in D$ we have $|f(y)|=0$ and hence $f(y)=0$.
answered Dec 16 '18 at 23:16
Wesley StrikWesley Strik
2,017423
$endgroup$
Yes it is correct, alternatively, as suggested by Quiliup, We can immediately conclude that since for all $y in D$ $$|f(y)|leq 0$$
By positive definiteness of the Euclidean norm it must hold that for all $y in D$ we have $|f(y)|=0$ and hence $f(y)=0$.
answered Dec 16 '18 at 23:16
Wesley StrikWesley Strik
2,017423
answered Dec 16 '18 at 23:16
Wesley StrikWesley Strik
2,017423
answered Dec 16 '18 at 23:16
Wesley StrikWesley Strik
2,017423
answered Dec 16 '18 at 23:16
Wesley StrikWesley Strik
2,017423
2,017423
add a comment |
add a comment |
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$begingroup$
It's correct, but for a proof this simple I would want to see the dummy variable that is being dropped from the argument: If $Vert fVert_{infty}=0$ then $0=Vert fVert_{infty}=sup_{x}|f(x)|geq|f(y)|$ for any $y$.
$endgroup$
– parsiad
Dec 16 '18 at 21:38
1
$begingroup$
This is Quite OK.
$endgroup$
– szw1710
Dec 16 '18 at 21:38
$begingroup$
Is this what you meant? I think that's a useful addition parsiad.
$endgroup$
– Wesley Strik
Dec 16 '18 at 21:40