Are the set of even functions a subset of continuous functions $C(-infty,+infty)$?
Multi tool use
$begingroup$
My book says that the above set is closed under scalar multiplication and addition. I can't seem to under why the given set is closed under scalar multiplication.
For instance, if I take an even function
$u=x^2$
Now check for scalar multiplication:
$cu=c(x^2)$ if $c<0$ (like $-x^2$)
Isn't closure under scalar multiplication now fail because $-x^2$ is not an even function?
linear-algebra proof-explanation
edited Jan 22 at 7:46
Robert Z
98.8k1068139
asked Jan 22 at 7:42
JohnySmith12JohnySmith12
393
$endgroup$
add a comment |
$begingroup$
My book says that the above set is closed under scalar multiplication and addition. I can't seem to under why the given set is closed under scalar multiplication.
For instance, if I take an even function
$u=x^2$
Now check for scalar multiplication:
$cu=c(x^2)$ if $c<0$ (like $-x^2$)
Isn't closure under scalar multiplication now fail because $-x^2$ is not an even function?
linear-algebra proof-explanation
edited Jan 22 at 7:46
Robert Z
98.8k1068139
asked Jan 22 at 7:42
JohnySmith12JohnySmith12
393
$endgroup$
$begingroup$
By "subset" did you mean "subspace"?
$endgroup$
– bof
Jan 22 at 7:52
$begingroup$
Also note that not all even functions are continuous.
$endgroup$
– gandalf61
Jan 22 at 10:47
add a comment |
$begingroup$
My book says that the above set is closed under scalar multiplication and addition. I can't seem to under why the given set is closed under scalar multiplication.
For instance, if I take an even function
$u=x^2$
Now check for scalar multiplication:
$cu=c(x^2)$ if $c<0$ (like $-x^2$)
Isn't closure under scalar multiplication now fail because $-x^2$ is not an even function?
linear-algebra proof-explanation
edited Jan 22 at 7:46
Robert Z
98.8k1068139
asked Jan 22 at 7:42
JohnySmith12JohnySmith12
393
$endgroup$
My book says that the above set is closed under scalar multiplication and addition. I can't seem to under why the given set is closed under scalar multiplication.
For instance, if I take an even function
$u=x^2$
Now check for scalar multiplication:
$cu=c(x^2)$ if $c<0$ (like $-x^2$)
Isn't closure under scalar multiplication now fail because $-x^2$ is not an even function?
linear-algebra proof-explanation
linear-algebra proof-explanation
edited Jan 22 at 7:46
Robert Z
98.8k1068139
asked Jan 22 at 7:42
JohnySmith12JohnySmith12
393
edited Jan 22 at 7:46
Robert Z
98.8k1068139
asked Jan 22 at 7:42
JohnySmith12JohnySmith12
393
edited Jan 22 at 7:46
Robert Z
98.8k1068139
edited Jan 22 at 7:46
Robert Z
98.8k1068139
edited Jan 22 at 7:46
Robert Z
98.8k1068139
98.8k1068139
asked Jan 22 at 7:42
JohnySmith12JohnySmith12
393
asked Jan 22 at 7:42
JohnySmith12JohnySmith12
393
asked Jan 22 at 7:42
JohnySmith12JohnySmith12
393
393
$begingroup$
By "subset" did you mean "subspace"?
$endgroup$
– bof
Jan 22 at 7:52
$begingroup$
Also note that not all even functions are continuous.
$endgroup$
– gandalf61
Jan 22 at 10:47
add a comment |
$begingroup$
By "subset" did you mean "subspace"?
$endgroup$
– bof
Jan 22 at 7:52
$begingroup$
Also note that not all even functions are continuous.
$endgroup$
– gandalf61
Jan 22 at 10:47
$begingroup$
By "subset" did you mean "subspace"?
$endgroup$
– bof
Jan 22 at 7:52
$begingroup$
By "subset" did you mean "subspace"?
$endgroup$
– bof
Jan 22 at 7:52
$begingroup$
Also note that not all even functions are continuous.
$endgroup$
– gandalf61
Jan 22 at 10:47
$begingroup$
Also note that not all even functions are continuous.
$endgroup$
– gandalf61
Jan 22 at 10:47
add a comment |
1 Answer
1
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$begingroup$
Yes, $f(x)=-x^2$ is an even function because it satisfies the definition:
$$f(-x)=-(-x)^2=-x^2=f(x)quadtext{ for all $x$ in the domain of $f$.}$$
In the same way you can verify that the set of even functions in $C(-infty,+infty)$ is closed under scalar multiplication.
answered Jan 22 at 7:44
Robert ZRobert Z
98.8k1068139
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, $f(x)=-x^2$ is an even function because it satisfies the definition:
$$f(-x)=-(-x)^2=-x^2=f(x)quadtext{ for all $x$ in the domain of $f$.}$$
In the same way you can verify that the set of even functions in $C(-infty,+infty)$ is closed under scalar multiplication.
answered Jan 22 at 7:44
Robert ZRobert Z
98.8k1068139
$endgroup$
add a comment |
$begingroup$
Yes, $f(x)=-x^2$ is an even function because it satisfies the definition:
$$f(-x)=-(-x)^2=-x^2=f(x)quadtext{ for all $x$ in the domain of $f$.}$$
In the same way you can verify that the set of even functions in $C(-infty,+infty)$ is closed under scalar multiplication.
answered Jan 22 at 7:44
Robert ZRobert Z
98.8k1068139
$endgroup$
add a comment |
$begingroup$
Yes, $f(x)=-x^2$ is an even function because it satisfies the definition:
$$f(-x)=-(-x)^2=-x^2=f(x)quadtext{ for all $x$ in the domain of $f$.}$$
In the same way you can verify that the set of even functions in $C(-infty,+infty)$ is closed under scalar multiplication.
answered Jan 22 at 7:44
Robert ZRobert Z
98.8k1068139
$endgroup$
Yes, $f(x)=-x^2$ is an even function because it satisfies the definition:
$$f(-x)=-(-x)^2=-x^2=f(x)quadtext{ for all $x$ in the domain of $f$.}$$
In the same way you can verify that the set of even functions in $C(-infty,+infty)$ is closed under scalar multiplication.
answered Jan 22 at 7:44
Robert ZRobert Z
98.8k1068139
edited Jan 22 at 7:50
answered Jan 22 at 7:44
Robert ZRobert Z
98.8k1068139
answered Jan 22 at 7:44
Robert ZRobert Z
98.8k1068139
answered Jan 22 at 7:44
Robert ZRobert Z
98.8k1068139
98.8k1068139
add a comment |
add a comment |
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P6kHyrgttgVLt6V6F41DbRie8gayvsba3m,SZcPrCxUsssgy,F4ej5p7W,nvz1jIFFHD
$begingroup$
By "subset" did you mean "subspace"?
$endgroup$
– bof
Jan 22 at 7:52
$begingroup$
Also note that not all even functions are continuous.
$endgroup$
– gandalf61
Jan 22 at 10:47