Continuity of function $T: left( C[0,1],d_{text{sup}} right) rightarrow left( C[0,1],d_{text{sup}} right)$
$begingroup$
Let $$T: left( C[0,1],d_{text{sup}} right) rightarrow left( C[0,1],d_{text{sup}} right)$$
where $$d_{text{sup}}(f,g) = mbox{sup}_{x in [0,1]} |f(x) - g(x)|$$
and the definition of $T$ is: $$T(f)(x) = 2 cdot f(1-x) - 3$$
Is it true that $T$ is a continuous function?
For me it is really strange example of function. Usually I consider function from real number to real number. But here I have no idea how to begin.
I will grateful for your help.
general-topology continuity
edited Nov 17 '13 at 13:14
Henno Brandsma
115k349125
asked Nov 17 '13 at 13:00
ThomasThomas
1,24411022
$endgroup$
|
show 3 more comments
$begingroup$
Let $$T: left( C[0,1],d_{text{sup}} right) rightarrow left( C[0,1],d_{text{sup}} right)$$
where $$d_{text{sup}}(f,g) = mbox{sup}_{x in [0,1]} |f(x) - g(x)|$$
and the definition of $T$ is: $$T(f)(x) = 2 cdot f(1-x) - 3$$
Is it true that $T$ is a continuous function?
For me it is really strange example of function. Usually I consider function from real number to real number. But here I have no idea how to begin.
I will grateful for your help.
general-topology continuity
edited Nov 17 '13 at 13:14
Henno Brandsma
115k349125
asked Nov 17 '13 at 13:00
ThomasThomas
1,24411022
$endgroup$
$begingroup$
what is the definition of $T$ here?
$endgroup$
– GA316
Nov 17 '13 at 13:02
$begingroup$
I added definition of $T$.
$endgroup$
– Thomas
Nov 17 '13 at 13:02
$begingroup$
Write it as a composition of simpler functions, $T = C circ B circ A$, where $A(f)(x) = f(1-x)$; $B(f) = 2cdot f$, $C(f)(x) = f(x)-3$. Check that $A,B,C$ are continuous.
$endgroup$
– Daniel Fischer
Nov 17 '13 at 13:23
$begingroup$
For me the problem is the same - how to check that $A,B,C$ are continuous?
$endgroup$
– Thomas
Nov 17 '13 at 13:24
$begingroup$
@DanielFischer You check that $T(f)$ is continuous, not $T$! Both need to be done.
$endgroup$
– Henno Brandsma
Nov 17 '13 at 13:37
|
show 3 more comments
$begingroup$
Let $$T: left( C[0,1],d_{text{sup}} right) rightarrow left( C[0,1],d_{text{sup}} right)$$
where $$d_{text{sup}}(f,g) = mbox{sup}_{x in [0,1]} |f(x) - g(x)|$$
and the definition of $T$ is: $$T(f)(x) = 2 cdot f(1-x) - 3$$
Is it true that $T$ is a continuous function?
For me it is really strange example of function. Usually I consider function from real number to real number. But here I have no idea how to begin.
I will grateful for your help.
general-topology continuity
edited Nov 17 '13 at 13:14
Henno Brandsma
115k349125
asked Nov 17 '13 at 13:00
ThomasThomas
1,24411022
$endgroup$
Let $$T: left( C[0,1],d_{text{sup}} right) rightarrow left( C[0,1],d_{text{sup}} right)$$
where $$d_{text{sup}}(f,g) = mbox{sup}_{x in [0,1]} |f(x) - g(x)|$$
and the definition of $T$ is: $$T(f)(x) = 2 cdot f(1-x) - 3$$
Is it true that $T$ is a continuous function?
For me it is really strange example of function. Usually I consider function from real number to real number. But here I have no idea how to begin.
I will grateful for your help.
general-topology continuity
general-topology continuity
edited Nov 17 '13 at 13:14
Henno Brandsma
115k349125
asked Nov 17 '13 at 13:00
ThomasThomas
1,24411022
edited Nov 17 '13 at 13:14
Henno Brandsma
115k349125
asked Nov 17 '13 at 13:00
ThomasThomas
1,24411022
edited Nov 17 '13 at 13:14
Henno Brandsma
115k349125
edited Nov 17 '13 at 13:14
Henno Brandsma
115k349125
edited Nov 17 '13 at 13:14
Henno Brandsma
115k349125
115k349125
asked Nov 17 '13 at 13:00
ThomasThomas
1,24411022
asked Nov 17 '13 at 13:00
ThomasThomas
1,24411022
asked Nov 17 '13 at 13:00
ThomasThomas
1,24411022
1,24411022
$begingroup$
what is the definition of $T$ here?
$endgroup$
– GA316
Nov 17 '13 at 13:02
$begingroup$
I added definition of $T$.
$endgroup$
– Thomas
Nov 17 '13 at 13:02
$begingroup$
Write it as a composition of simpler functions, $T = C circ B circ A$, where $A(f)(x) = f(1-x)$; $B(f) = 2cdot f$, $C(f)(x) = f(x)-3$. Check that $A,B,C$ are continuous.
$endgroup$
– Daniel Fischer
Nov 17 '13 at 13:23
$begingroup$
For me the problem is the same - how to check that $A,B,C$ are continuous?
$endgroup$
– Thomas
Nov 17 '13 at 13:24
$begingroup$
@DanielFischer You check that $T(f)$ is continuous, not $T$! Both need to be done.
$endgroup$
– Henno Brandsma
Nov 17 '13 at 13:37
|
show 3 more comments
$begingroup$
what is the definition of $T$ here?
$endgroup$
– GA316
Nov 17 '13 at 13:02
$begingroup$
I added definition of $T$.
$endgroup$
– Thomas
Nov 17 '13 at 13:02
$begingroup$
Write it as a composition of simpler functions, $T = C circ B circ A$, where $A(f)(x) = f(1-x)$; $B(f) = 2cdot f$, $C(f)(x) = f(x)-3$. Check that $A,B,C$ are continuous.
$endgroup$
– Daniel Fischer
Nov 17 '13 at 13:23
$begingroup$
For me the problem is the same - how to check that $A,B,C$ are continuous?
$endgroup$
– Thomas
Nov 17 '13 at 13:24
$begingroup$
@DanielFischer You check that $T(f)$ is continuous, not $T$! Both need to be done.
$endgroup$
– Henno Brandsma
Nov 17 '13 at 13:37
$begingroup$
what is the definition of $T$ here?
$endgroup$
– GA316
Nov 17 '13 at 13:02
$begingroup$
what is the definition of $T$ here?
$endgroup$
– GA316
Nov 17 '13 at 13:02
$begingroup$
I added definition of $T$.
$endgroup$
– Thomas
Nov 17 '13 at 13:02
$begingroup$
I added definition of $T$.
$endgroup$
– Thomas
Nov 17 '13 at 13:02
$begingroup$
Write it as a composition of simpler functions, $T = C circ B circ A$, where $A(f)(x) = f(1-x)$; $B(f) = 2cdot f$, $C(f)(x) = f(x)-3$. Check that $A,B,C$ are continuous.
$endgroup$
– Daniel Fischer
Nov 17 '13 at 13:23
$begingroup$
Write it as a composition of simpler functions, $T = C circ B circ A$, where $A(f)(x) = f(1-x)$; $B(f) = 2cdot f$, $C(f)(x) = f(x)-3$. Check that $A,B,C$ are continuous.
$endgroup$
– Daniel Fischer
Nov 17 '13 at 13:23
$begingroup$
For me the problem is the same - how to check that $A,B,C$ are continuous?
$endgroup$
– Thomas
Nov 17 '13 at 13:24
$begingroup$
For me the problem is the same - how to check that $A,B,C$ are continuous?
$endgroup$
– Thomas
Nov 17 '13 at 13:24
$begingroup$
@DanielFischer You check that $T(f)$ is continuous, not $T$! Both need to be done.
$endgroup$
– Henno Brandsma
Nov 17 '13 at 13:37
$begingroup$
@DanielFischer You check that $T(f)$ is continuous, not $T$! Both need to be done.
$endgroup$
– Henno Brandsma
Nov 17 '13 at 13:37
|
show 3 more comments
1 Answer
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$begingroup$
For any set $S$ it makes sense to talk about a function $f: S rightarrow S$, formally this is a set of ordered pairs satisfying some conditions, but a more informal description is that we need for every $s$ in $S$, some unique value $f(s)$ in $S$ as well. In the case at hand, $S$ is a set of (continuous) functions, not a set of numbers, but the principle is the same.
So $T(f)$ must be in the set as well, whenever $f$ is, and you described the rule: as $T(f)$ is itself a function from $[0,1]$ to $mathbb{R}$, to describe what it is we need to define what $T(f)(x)$ is for every $x$ in $[0,1]$, and it takes the value of $f$ (the input function) at the point $1-x$, multiplies it by 2 and subtracts 3: this is again some real number, uniquely determined when we know $f$, so $T(f)(x)$ is well-defined for all $x in [0,1]$. Whenever $f$ is continuous between $[0,1]$ and $mathbb{R}$, so is $T(f)$: we can see $T(f)(x)$ as a composition of continuous functions ($x rightarrow 1-x, f, t rightarrow 2t-3$, informally), and so $f$ maps continuous functions to continuous functions as required.
To see continuity in the metric $d_sup$, you need to compute what $d_sup(T(f), T(g))$ is, for functions $f,g$. By definition this equals the sup (actually a maximum as we work with
continuous functions on a compact space) of all differences $left|T(f)(x) - T(g)(x)right|$
where $x$ runs over all values in $[0,1]$. But this equals (using definitions and cancelling the $-3$'s) $$|,2cdot f(1-x) - 2 cdot g(1-x) ,| = 2cdot|,f(1-x) - g(1-x),|$$
and the sup of these over all $x$ is just the same as the sup of $|f(x)-g(x)|$ over all $x$ : if one assumes its max at $a$, the other assumes it at $1-a$ and vice versa).
In short, $d_sup(T(f), T(g)) = 2d_sup(f,g)$, and this will easily imply continuity of $T$..
answered Nov 17 '13 at 13:36
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
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$begingroup$
For any set $S$ it makes sense to talk about a function $f: S rightarrow S$, formally this is a set of ordered pairs satisfying some conditions, but a more informal description is that we need for every $s$ in $S$, some unique value $f(s)$ in $S$ as well. In the case at hand, $S$ is a set of (continuous) functions, not a set of numbers, but the principle is the same.
So $T(f)$ must be in the set as well, whenever $f$ is, and you described the rule: as $T(f)$ is itself a function from $[0,1]$ to $mathbb{R}$, to describe what it is we need to define what $T(f)(x)$ is for every $x$ in $[0,1]$, and it takes the value of $f$ (the input function) at the point $1-x$, multiplies it by 2 and subtracts 3: this is again some real number, uniquely determined when we know $f$, so $T(f)(x)$ is well-defined for all $x in [0,1]$. Whenever $f$ is continuous between $[0,1]$ and $mathbb{R}$, so is $T(f)$: we can see $T(f)(x)$ as a composition of continuous functions ($x rightarrow 1-x, f, t rightarrow 2t-3$, informally), and so $f$ maps continuous functions to continuous functions as required.
To see continuity in the metric $d_sup$, you need to compute what $d_sup(T(f), T(g))$ is, for functions $f,g$. By definition this equals the sup (actually a maximum as we work with
continuous functions on a compact space) of all differences $left|T(f)(x) - T(g)(x)right|$
where $x$ runs over all values in $[0,1]$. But this equals (using definitions and cancelling the $-3$'s) $$|,2cdot f(1-x) - 2 cdot g(1-x) ,| = 2cdot|,f(1-x) - g(1-x),|$$
and the sup of these over all $x$ is just the same as the sup of $|f(x)-g(x)|$ over all $x$ : if one assumes its max at $a$, the other assumes it at $1-a$ and vice versa).
In short, $d_sup(T(f), T(g)) = 2d_sup(f,g)$, and this will easily imply continuity of $T$..
answered Nov 17 '13 at 13:36
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
$begingroup$
For any set $S$ it makes sense to talk about a function $f: S rightarrow S$, formally this is a set of ordered pairs satisfying some conditions, but a more informal description is that we need for every $s$ in $S$, some unique value $f(s)$ in $S$ as well. In the case at hand, $S$ is a set of (continuous) functions, not a set of numbers, but the principle is the same.
So $T(f)$ must be in the set as well, whenever $f$ is, and you described the rule: as $T(f)$ is itself a function from $[0,1]$ to $mathbb{R}$, to describe what it is we need to define what $T(f)(x)$ is for every $x$ in $[0,1]$, and it takes the value of $f$ (the input function) at the point $1-x$, multiplies it by 2 and subtracts 3: this is again some real number, uniquely determined when we know $f$, so $T(f)(x)$ is well-defined for all $x in [0,1]$. Whenever $f$ is continuous between $[0,1]$ and $mathbb{R}$, so is $T(f)$: we can see $T(f)(x)$ as a composition of continuous functions ($x rightarrow 1-x, f, t rightarrow 2t-3$, informally), and so $f$ maps continuous functions to continuous functions as required.
To see continuity in the metric $d_sup$, you need to compute what $d_sup(T(f), T(g))$ is, for functions $f,g$. By definition this equals the sup (actually a maximum as we work with
continuous functions on a compact space) of all differences $left|T(f)(x) - T(g)(x)right|$
where $x$ runs over all values in $[0,1]$. But this equals (using definitions and cancelling the $-3$'s) $$|,2cdot f(1-x) - 2 cdot g(1-x) ,| = 2cdot|,f(1-x) - g(1-x),|$$
and the sup of these over all $x$ is just the same as the sup of $|f(x)-g(x)|$ over all $x$ : if one assumes its max at $a$, the other assumes it at $1-a$ and vice versa).
In short, $d_sup(T(f), T(g)) = 2d_sup(f,g)$, and this will easily imply continuity of $T$..
answered Nov 17 '13 at 13:36
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
$begingroup$
For any set $S$ it makes sense to talk about a function $f: S rightarrow S$, formally this is a set of ordered pairs satisfying some conditions, but a more informal description is that we need for every $s$ in $S$, some unique value $f(s)$ in $S$ as well. In the case at hand, $S$ is a set of (continuous) functions, not a set of numbers, but the principle is the same.
So $T(f)$ must be in the set as well, whenever $f$ is, and you described the rule: as $T(f)$ is itself a function from $[0,1]$ to $mathbb{R}$, to describe what it is we need to define what $T(f)(x)$ is for every $x$ in $[0,1]$, and it takes the value of $f$ (the input function) at the point $1-x$, multiplies it by 2 and subtracts 3: this is again some real number, uniquely determined when we know $f$, so $T(f)(x)$ is well-defined for all $x in [0,1]$. Whenever $f$ is continuous between $[0,1]$ and $mathbb{R}$, so is $T(f)$: we can see $T(f)(x)$ as a composition of continuous functions ($x rightarrow 1-x, f, t rightarrow 2t-3$, informally), and so $f$ maps continuous functions to continuous functions as required.
To see continuity in the metric $d_sup$, you need to compute what $d_sup(T(f), T(g))$ is, for functions $f,g$. By definition this equals the sup (actually a maximum as we work with
continuous functions on a compact space) of all differences $left|T(f)(x) - T(g)(x)right|$
where $x$ runs over all values in $[0,1]$. But this equals (using definitions and cancelling the $-3$'s) $$|,2cdot f(1-x) - 2 cdot g(1-x) ,| = 2cdot|,f(1-x) - g(1-x),|$$
and the sup of these over all $x$ is just the same as the sup of $|f(x)-g(x)|$ over all $x$ : if one assumes its max at $a$, the other assumes it at $1-a$ and vice versa).
In short, $d_sup(T(f), T(g)) = 2d_sup(f,g)$, and this will easily imply continuity of $T$..
answered Nov 17 '13 at 13:36
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
For any set $S$ it makes sense to talk about a function $f: S rightarrow S$, formally this is a set of ordered pairs satisfying some conditions, but a more informal description is that we need for every $s$ in $S$, some unique value $f(s)$ in $S$ as well. In the case at hand, $S$ is a set of (continuous) functions, not a set of numbers, but the principle is the same.
So $T(f)$ must be in the set as well, whenever $f$ is, and you described the rule: as $T(f)$ is itself a function from $[0,1]$ to $mathbb{R}$, to describe what it is we need to define what $T(f)(x)$ is for every $x$ in $[0,1]$, and it takes the value of $f$ (the input function) at the point $1-x$, multiplies it by 2 and subtracts 3: this is again some real number, uniquely determined when we know $f$, so $T(f)(x)$ is well-defined for all $x in [0,1]$. Whenever $f$ is continuous between $[0,1]$ and $mathbb{R}$, so is $T(f)$: we can see $T(f)(x)$ as a composition of continuous functions ($x rightarrow 1-x, f, t rightarrow 2t-3$, informally), and so $f$ maps continuous functions to continuous functions as required.
To see continuity in the metric $d_sup$, you need to compute what $d_sup(T(f), T(g))$ is, for functions $f,g$. By definition this equals the sup (actually a maximum as we work with
continuous functions on a compact space) of all differences $left|T(f)(x) - T(g)(x)right|$
where $x$ runs over all values in $[0,1]$. But this equals (using definitions and cancelling the $-3$'s) $$|,2cdot f(1-x) - 2 cdot g(1-x) ,| = 2cdot|,f(1-x) - g(1-x),|$$
and the sup of these over all $x$ is just the same as the sup of $|f(x)-g(x)|$ over all $x$ : if one assumes its max at $a$, the other assumes it at $1-a$ and vice versa).
In short, $d_sup(T(f), T(g)) = 2d_sup(f,g)$, and this will easily imply continuity of $T$..
answered Nov 17 '13 at 13:36
Henno BrandsmaHenno Brandsma
115k349125
edited Jan 3 at 18:00
answered Nov 17 '13 at 13:36
Henno BrandsmaHenno Brandsma
115k349125
answered Nov 17 '13 at 13:36
Henno BrandsmaHenno Brandsma
115k349125
answered Nov 17 '13 at 13:36
Henno BrandsmaHenno Brandsma
115k349125
115k349125
add a comment |
add a comment |
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$begingroup$
what is the definition of $T$ here?
$endgroup$
– GA316
Nov 17 '13 at 13:02
$begingroup$
I added definition of $T$.
$endgroup$
– Thomas
Nov 17 '13 at 13:02
$begingroup$
Write it as a composition of simpler functions, $T = C circ B circ A$, where $A(f)(x) = f(1-x)$; $B(f) = 2cdot f$, $C(f)(x) = f(x)-3$. Check that $A,B,C$ are continuous.
$endgroup$
– Daniel Fischer
Nov 17 '13 at 13:23
$begingroup$
For me the problem is the same - how to check that $A,B,C$ are continuous?
$endgroup$
– Thomas
Nov 17 '13 at 13:24
$begingroup$
@DanielFischer You check that $T(f)$ is continuous, not $T$! Both need to be done.
$endgroup$
– Henno Brandsma
Nov 17 '13 at 13:37