$f:[0,1] times [0,1] → mathbb R$ is continuous. Prove that $g(x) = max{f(x,y) : y in [0,1]}$ is defined and...
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Let $f:[0,1] times [0,1] → mathbb R$ be a continuous function. Prove that $$g(x) = max{f(x,y) : y in [0,1]}$$ makes sense (in that the maximum exists) and is continuous.
I said that for each $x$, we can consider $A_x = {x} times [0,1] subset text{dom } f $. $A_x$ is compact, and since $max$, which is acting on a continuous function $f$, is itself continuous,we know that (a) $g$ is continuous, and (b) the image $g(A_x)$ is compact, which means it attains a maximum and minimum. Therefore the maximum exists for each $x$, and by (a) $g$ is continuous.
Edit: It seems that the fact that $A_x$ isn't fixed makes the proof fail. What would a working proof look like?
real-analysis proof-verification continuity
asked Nov 14 at 10:57
Tiwa Aina
2,671320
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up vote
2
down vote
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Let $f:[0,1] times [0,1] → mathbb R$ be a continuous function. Prove that $$g(x) = max{f(x,y) : y in [0,1]}$$ makes sense (in that the maximum exists) and is continuous.
I said that for each $x$, we can consider $A_x = {x} times [0,1] subset text{dom } f $. $A_x$ is compact, and since $max$, which is acting on a continuous function $f$, is itself continuous,we know that (a) $g$ is continuous, and (b) the image $g(A_x)$ is compact, which means it attains a maximum and minimum. Therefore the maximum exists for each $x$, and by (a) $g$ is continuous.
Edit: It seems that the fact that $A_x$ isn't fixed makes the proof fail. What would a working proof look like?
real-analysis proof-verification continuity
asked Nov 14 at 10:57
Tiwa Aina
2,671320
No, this reasoning is not entirely valid since $A$ is not a fixed compact set (it varies with $x$ ; you should denote it by $A(x)$ or $A_x$). So your argument does show that $g$ is well-defined but not that $g$ is continuous.
– Ewan Delanoy
Nov 14 at 11:01
@EwanDelanoy Fixed the notation. What part of the proof does $A$'s non-constancy cause to fail?
– Tiwa Aina
Nov 14 at 11:03
I already answered that in my former comment : it's the part that says that $g$ is continuous. $g$ is not defined on $A_x$, it is defined on $[0,1]$. $f$ is defined on all of $[0,1]^2$, and also by restriction on $A_x$. Note that $f$ is a two-variables function while $g$ is a one-variable function.
– Ewan Delanoy
Nov 14 at 12:10
2
Uniform continuity of $f$ can be shown to imply continuity of $g$.
– random
Nov 14 at 12:34
This question is relevant (math.stackexchange.com/questions/2740283/…). You can show that the family ${f_x}$ is equicontinuous (using uniform continuity of $f$), and note $g$ is the supremum of ${f_x}$
– user25959
Nov 14 at 14:52
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f:[0,1] times [0,1] → mathbb R$ be a continuous function. Prove that $$g(x) = max{f(x,y) : y in [0,1]}$$ makes sense (in that the maximum exists) and is continuous.
I said that for each $x$, we can consider $A_x = {x} times [0,1] subset text{dom } f $. $A_x$ is compact, and since $max$, which is acting on a continuous function $f$, is itself continuous,we know that (a) $g$ is continuous, and (b) the image $g(A_x)$ is compact, which means it attains a maximum and minimum. Therefore the maximum exists for each $x$, and by (a) $g$ is continuous.
Edit: It seems that the fact that $A_x$ isn't fixed makes the proof fail. What would a working proof look like?
real-analysis proof-verification continuity
asked Nov 14 at 10:57
Tiwa Aina
2,671320
Let $f:[0,1] times [0,1] → mathbb R$ be a continuous function. Prove that $$g(x) = max{f(x,y) : y in [0,1]}$$ makes sense (in that the maximum exists) and is continuous.
I said that for each $x$, we can consider $A_x = {x} times [0,1] subset text{dom } f $. $A_x$ is compact, and since $max$, which is acting on a continuous function $f$, is itself continuous,we know that (a) $g$ is continuous, and (b) the image $g(A_x)$ is compact, which means it attains a maximum and minimum. Therefore the maximum exists for each $x$, and by (a) $g$ is continuous.
Edit: It seems that the fact that $A_x$ isn't fixed makes the proof fail. What would a working proof look like?
real-analysis proof-verification continuity
real-analysis proof-verification continuity
asked Nov 14 at 10:57
Tiwa Aina
2,671320
asked Nov 14 at 10:57
Tiwa Aina
2,671320
edited Nov 14 at 14:48
asked Nov 14 at 10:57
Tiwa Aina
2,671320
asked Nov 14 at 10:57
Tiwa Aina
2,671320
asked Nov 14 at 10:57
Tiwa Aina
2,671320
2,671320
No, this reasoning is not entirely valid since $A$ is not a fixed compact set (it varies with $x$ ; you should denote it by $A(x)$ or $A_x$). So your argument does show that $g$ is well-defined but not that $g$ is continuous.
– Ewan Delanoy
Nov 14 at 11:01
@EwanDelanoy Fixed the notation. What part of the proof does $A$'s non-constancy cause to fail?
– Tiwa Aina
Nov 14 at 11:03
I already answered that in my former comment : it's the part that says that $g$ is continuous. $g$ is not defined on $A_x$, it is defined on $[0,1]$. $f$ is defined on all of $[0,1]^2$, and also by restriction on $A_x$. Note that $f$ is a two-variables function while $g$ is a one-variable function.
– Ewan Delanoy
Nov 14 at 12:10
2
Uniform continuity of $f$ can be shown to imply continuity of $g$.
– random
Nov 14 at 12:34
This question is relevant (math.stackexchange.com/questions/2740283/…). You can show that the family ${f_x}$ is equicontinuous (using uniform continuity of $f$), and note $g$ is the supremum of ${f_x}$
– user25959
Nov 14 at 14:52
add a comment |
No, this reasoning is not entirely valid since $A$ is not a fixed compact set (it varies with $x$ ; you should denote it by $A(x)$ or $A_x$). So your argument does show that $g$ is well-defined but not that $g$ is continuous.
– Ewan Delanoy
Nov 14 at 11:01
@EwanDelanoy Fixed the notation. What part of the proof does $A$'s non-constancy cause to fail?
– Tiwa Aina
Nov 14 at 11:03
I already answered that in my former comment : it's the part that says that $g$ is continuous. $g$ is not defined on $A_x$, it is defined on $[0,1]$. $f$ is defined on all of $[0,1]^2$, and also by restriction on $A_x$. Note that $f$ is a two-variables function while $g$ is a one-variable function.
– Ewan Delanoy
Nov 14 at 12:10
2
Uniform continuity of $f$ can be shown to imply continuity of $g$.
– random
Nov 14 at 12:34
This question is relevant (math.stackexchange.com/questions/2740283/…). You can show that the family ${f_x}$ is equicontinuous (using uniform continuity of $f$), and note $g$ is the supremum of ${f_x}$
– user25959
Nov 14 at 14:52
No, this reasoning is not entirely valid since $A$ is not a fixed compact set (it varies with $x$ ; you should denote it by $A(x)$ or $A_x$). So your argument does show that $g$ is well-defined but not that $g$ is continuous.
– Ewan Delanoy
Nov 14 at 11:01
No, this reasoning is not entirely valid since $A$ is not a fixed compact set (it varies with $x$ ; you should denote it by $A(x)$ or $A_x$). So your argument does show that $g$ is well-defined but not that $g$ is continuous.
– Ewan Delanoy
Nov 14 at 11:01
@EwanDelanoy Fixed the notation. What part of the proof does $A$'s non-constancy cause to fail?
– Tiwa Aina
Nov 14 at 11:03
@EwanDelanoy Fixed the notation. What part of the proof does $A$'s non-constancy cause to fail?
– Tiwa Aina
Nov 14 at 11:03
I already answered that in my former comment : it's the part that says that $g$ is continuous. $g$ is not defined on $A_x$, it is defined on $[0,1]$. $f$ is defined on all of $[0,1]^2$, and also by restriction on $A_x$. Note that $f$ is a two-variables function while $g$ is a one-variable function.
– Ewan Delanoy
Nov 14 at 12:10
I already answered that in my former comment : it's the part that says that $g$ is continuous. $g$ is not defined on $A_x$, it is defined on $[0,1]$. $f$ is defined on all of $[0,1]^2$, and also by restriction on $A_x$. Note that $f$ is a two-variables function while $g$ is a one-variable function.
– Ewan Delanoy
Nov 14 at 12:10
2
2
Uniform continuity of $f$ can be shown to imply continuity of $g$.
– random
Nov 14 at 12:34
Uniform continuity of $f$ can be shown to imply continuity of $g$.
– random
Nov 14 at 12:34
This question is relevant (math.stackexchange.com/questions/2740283/…). You can show that the family ${f_x}$ is equicontinuous (using uniform continuity of $f$), and note $g$ is the supremum of ${f_x}$
– user25959
Nov 14 at 14:52
This question is relevant (math.stackexchange.com/questions/2740283/…). You can show that the family ${f_x}$ is equicontinuous (using uniform continuity of $f$), and note $g$ is the supremum of ${f_x}$
– user25959
Nov 14 at 14:52
add a comment |
1 Answer
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The continuity with respect to $y$ is irrelevant as long as the functions $xmapsto f(x,y)$ are equicontinuous with respect to $y$: Given $x_0$ and an $epsilon>0$, there is a $delta>0$ such that $$|x-x_0|<deltaqquadRightarrowqquad|f(x,y)-f(x_0,y)|<epsilonquadforall y .tag{1}$$
Let an $x_0in[0,1]$ and an $epsilon>0$ be given. Choose a $delta>0$ such that $(1)$ holds. If $|x-x_0|<delta$ then
$$f(x,y)leq f(x_0,y)+epsilon leq g(x_0)+epsilon qquadforall y .$$
It follows that $g(x)leq g(x_0)+epsilon$. Similarly
$$f(x_0,y)leq f(x,y)+epsilonleq g(x)+epsilonquadforall y ,$$
and therefore $g(x_0)leq g(x)+epsilon$. In all we have proven that $|x-x_0|<delta$ implies $|g(x)-g(x_0)|leqepsilon$, as required.
answered Nov 14 at 16:07
Christian Blatter
170k7111324
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The continuity with respect to $y$ is irrelevant as long as the functions $xmapsto f(x,y)$ are equicontinuous with respect to $y$: Given $x_0$ and an $epsilon>0$, there is a $delta>0$ such that $$|x-x_0|<deltaqquadRightarrowqquad|f(x,y)-f(x_0,y)|<epsilonquadforall y .tag{1}$$
Let an $x_0in[0,1]$ and an $epsilon>0$ be given. Choose a $delta>0$ such that $(1)$ holds. If $|x-x_0|<delta$ then
$$f(x,y)leq f(x_0,y)+epsilon leq g(x_0)+epsilon qquadforall y .$$
It follows that $g(x)leq g(x_0)+epsilon$. Similarly
$$f(x_0,y)leq f(x,y)+epsilonleq g(x)+epsilonquadforall y ,$$
and therefore $g(x_0)leq g(x)+epsilon$. In all we have proven that $|x-x_0|<delta$ implies $|g(x)-g(x_0)|leqepsilon$, as required.
answered Nov 14 at 16:07
Christian Blatter
170k7111324
add a comment |
up vote
1
down vote
accepted
The continuity with respect to $y$ is irrelevant as long as the functions $xmapsto f(x,y)$ are equicontinuous with respect to $y$: Given $x_0$ and an $epsilon>0$, there is a $delta>0$ such that $$|x-x_0|<deltaqquadRightarrowqquad|f(x,y)-f(x_0,y)|<epsilonquadforall y .tag{1}$$
Let an $x_0in[0,1]$ and an $epsilon>0$ be given. Choose a $delta>0$ such that $(1)$ holds. If $|x-x_0|<delta$ then
$$f(x,y)leq f(x_0,y)+epsilon leq g(x_0)+epsilon qquadforall y .$$
It follows that $g(x)leq g(x_0)+epsilon$. Similarly
$$f(x_0,y)leq f(x,y)+epsilonleq g(x)+epsilonquadforall y ,$$
and therefore $g(x_0)leq g(x)+epsilon$. In all we have proven that $|x-x_0|<delta$ implies $|g(x)-g(x_0)|leqepsilon$, as required.
answered Nov 14 at 16:07
Christian Blatter
170k7111324
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The continuity with respect to $y$ is irrelevant as long as the functions $xmapsto f(x,y)$ are equicontinuous with respect to $y$: Given $x_0$ and an $epsilon>0$, there is a $delta>0$ such that $$|x-x_0|<deltaqquadRightarrowqquad|f(x,y)-f(x_0,y)|<epsilonquadforall y .tag{1}$$
Let an $x_0in[0,1]$ and an $epsilon>0$ be given. Choose a $delta>0$ such that $(1)$ holds. If $|x-x_0|<delta$ then
$$f(x,y)leq f(x_0,y)+epsilon leq g(x_0)+epsilon qquadforall y .$$
It follows that $g(x)leq g(x_0)+epsilon$. Similarly
$$f(x_0,y)leq f(x,y)+epsilonleq g(x)+epsilonquadforall y ,$$
and therefore $g(x_0)leq g(x)+epsilon$. In all we have proven that $|x-x_0|<delta$ implies $|g(x)-g(x_0)|leqepsilon$, as required.
answered Nov 14 at 16:07
Christian Blatter
170k7111324
The continuity with respect to $y$ is irrelevant as long as the functions $xmapsto f(x,y)$ are equicontinuous with respect to $y$: Given $x_0$ and an $epsilon>0$, there is a $delta>0$ such that $$|x-x_0|<deltaqquadRightarrowqquad|f(x,y)-f(x_0,y)|<epsilonquadforall y .tag{1}$$
Let an $x_0in[0,1]$ and an $epsilon>0$ be given. Choose a $delta>0$ such that $(1)$ holds. If $|x-x_0|<delta$ then
$$f(x,y)leq f(x_0,y)+epsilon leq g(x_0)+epsilon qquadforall y .$$
It follows that $g(x)leq g(x_0)+epsilon$. Similarly
$$f(x_0,y)leq f(x,y)+epsilonleq g(x)+epsilonquadforall y ,$$
and therefore $g(x_0)leq g(x)+epsilon$. In all we have proven that $|x-x_0|<delta$ implies $|g(x)-g(x_0)|leqepsilon$, as required.
answered Nov 14 at 16:07
Christian Blatter
170k7111324
answered Nov 14 at 16:07
Christian Blatter
170k7111324
answered Nov 14 at 16:07
Christian Blatter
170k7111324
answered Nov 14 at 16:07
Christian Blatter
170k7111324
170k7111324
add a comment |
add a comment |
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No, this reasoning is not entirely valid since $A$ is not a fixed compact set (it varies with $x$ ; you should denote it by $A(x)$ or $A_x$). So your argument does show that $g$ is well-defined but not that $g$ is continuous.
– Ewan Delanoy
Nov 14 at 11:01
@EwanDelanoy Fixed the notation. What part of the proof does $A$'s non-constancy cause to fail?
– Tiwa Aina
Nov 14 at 11:03
I already answered that in my former comment : it's the part that says that $g$ is continuous. $g$ is not defined on $A_x$, it is defined on $[0,1]$. $f$ is defined on all of $[0,1]^2$, and also by restriction on $A_x$. Note that $f$ is a two-variables function while $g$ is a one-variable function.
– Ewan Delanoy
Nov 14 at 12:10
2
Uniform continuity of $f$ can be shown to imply continuity of $g$.
– random
Nov 14 at 12:34
This question is relevant (math.stackexchange.com/questions/2740283/…). You can show that the family ${f_x}$ is equicontinuous (using uniform continuity of $f$), and note $g$ is the supremum of ${f_x}$
– user25959
Nov 14 at 14:52