Continuity of function $T: left( C[0,1],d_{text{sup}} right) rightarrow left( C[0,1],d_{text{sup}} right)$












0












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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.










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 17 '13 at 13:14









Henno Brandsma

107k347114




107k347114










asked Nov 17 '13 at 13:00









ThomasThomas

1,23911021




1,23911021












  • $begingroup$
    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












  • $begingroup$
    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










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 $T$..






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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 $T$..






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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 $T$..






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $T$..






        share|cite|improve this answer











        $endgroup$



        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 $T$..







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 18:00

























        answered Nov 17 '13 at 13:36









        Henno BrandsmaHenno Brandsma

        107k347114




        107k347114






























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