Show that $ inf_{c in mathbb{R}} |f - c|_{infty} = | f - frac{1}{2}(min f + max f) |_{infty}$












1














I'm very unsure about what I did here, so if someone could proof check on the following, I would be very thankful.



Let $I$ be a an interval of $mathbb{R}$.



For $f in mathcal{C}(I)$, we say that $P in mathcal{P}_n$ ($P$ is a polynomial of degree $n$) is the best approximation to $f$ if $| f- P |_{infty} = inf , { |f-P|_{infty} : P in mathcal{P}_n } $.



I want to show that the best approximation of $f in mathcal{C}(I)$ by a constant is $frac{1}{2}(inf_{x in I} f(x) + sup_{x in I} f(x))$.



$text{Proof.}$



Since $f$ is continuous and $I$ is compact, $inf f$ and $sup f$ are reached and we have that $inf f < infty$ and $sup f < infty$.



Hence our claim is equivalent to saying that
$inf , { |f - c|_{infty}: c in mathbb{R}} = | f - frac{1}{2}(min f + max f) |_{infty}$



Note that we have that $inf_{c in mathbb{R}} |f - c |_{infty} leq |f - frac{1}{2}(min f + max f) |_{infty}$



Thus it suffices to show that for all $varepsilon > 0$, there exists $c in mathbb{R}$ such that $|f-c|_{infty} < | f - frac{1}{2}(min f + max f) |_{infty} + varepsilon$



Let $varepsilon > 0$. Define $c =: frac{1}{2}(min f + max f - varepsilon)$.
We have that
$$ |f - c|_{infty} = | f - frac{1}{2}(min f + max f) + frac{1}{2}(min f + max f) -c|_{infty} leq$$
$$ leq | f - frac{1}{2}(min f + max f)|_{infty} + | frac{1}{2}(min f + max f) -c |_{infty}= $$
$$ = | f - frac{1}{2}(min f + max f)|_{infty} + frac{1}{2}varepsilon <$$
$$ < | f - frac{1}{2}(min f + max f)|_{infty} + varepsilon$$






real-analysis functional-analysis proof-verification numerical-methods







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  • 1




    Closed and bounded does not imply interval without connectedness. More importantly, from your chain of inequalities, we only get back to $inf |f-c| le | f - ... | $, and the other direction is still needed ( $inf |f-c| ge | f - ... |$)
    – Calvin Khor
    Dec 8 at 22:23










  • could you give me a hint on how to show the other inequality ?
    – riri92
    Dec 8 at 22:29
















1














I'm very unsure about what I did here, so if someone could proof check on the following, I would be very thankful.



Let $I$ be a an interval of $mathbb{R}$.



For $f in mathcal{C}(I)$, we say that $P in mathcal{P}_n$ ($P$ is a polynomial of degree $n$) is the best approximation to $f$ if $| f- P |_{infty} = inf , { |f-P|_{infty} : P in mathcal{P}_n } $.



I want to show that the best approximation of $f in mathcal{C}(I)$ by a constant is $frac{1}{2}(inf_{x in I} f(x) + sup_{x in I} f(x))$.



$text{Proof.}$



Since $f$ is continuous and $I$ is compact, $inf f$ and $sup f$ are reached and we have that $inf f < infty$ and $sup f < infty$.



Hence our claim is equivalent to saying that
$inf , { |f - c|_{infty}: c in mathbb{R}} = | f - frac{1}{2}(min f + max f) |_{infty}$



Note that we have that $inf_{c in mathbb{R}} |f - c |_{infty} leq |f - frac{1}{2}(min f + max f) |_{infty}$



Thus it suffices to show that for all $varepsilon > 0$, there exists $c in mathbb{R}$ such that $|f-c|_{infty} < | f - frac{1}{2}(min f + max f) |_{infty} + varepsilon$



Let $varepsilon > 0$. Define $c =: frac{1}{2}(min f + max f - varepsilon)$.
We have that
$$ |f - c|_{infty} = | f - frac{1}{2}(min f + max f) + frac{1}{2}(min f + max f) -c|_{infty} leq$$
$$ leq | f - frac{1}{2}(min f + max f)|_{infty} + | frac{1}{2}(min f + max f) -c |_{infty}= $$
$$ = | f - frac{1}{2}(min f + max f)|_{infty} + frac{1}{2}varepsilon <$$
$$ < | f - frac{1}{2}(min f + max f)|_{infty} + varepsilon$$






real-analysis functional-analysis proof-verification numerical-methods







share|cite|improve this question




















  • 1




    Closed and bounded does not imply interval without connectedness. More importantly, from your chain of inequalities, we only get back to $inf |f-c| le | f - ... | $, and the other direction is still needed ( $inf |f-c| ge | f - ... |$)
    – Calvin Khor
    Dec 8 at 22:23










  • could you give me a hint on how to show the other inequality ?
    – riri92
    Dec 8 at 22:29














1












1








1


1





I'm very unsure about what I did here, so if someone could proof check on the following, I would be very thankful.



Let $I$ be a an interval of $mathbb{R}$.



For $f in mathcal{C}(I)$, we say that $P in mathcal{P}_n$ ($P$ is a polynomial of degree $n$) is the best approximation to $f$ if $| f- P |_{infty} = inf , { |f-P|_{infty} : P in mathcal{P}_n } $.



I want to show that the best approximation of $f in mathcal{C}(I)$ by a constant is $frac{1}{2}(inf_{x in I} f(x) + sup_{x in I} f(x))$.



$text{Proof.}$



Since $f$ is continuous and $I$ is compact, $inf f$ and $sup f$ are reached and we have that $inf f < infty$ and $sup f < infty$.



Hence our claim is equivalent to saying that
$inf , { |f - c|_{infty}: c in mathbb{R}} = | f - frac{1}{2}(min f + max f) |_{infty}$



Note that we have that $inf_{c in mathbb{R}} |f - c |_{infty} leq |f - frac{1}{2}(min f + max f) |_{infty}$



Thus it suffices to show that for all $varepsilon > 0$, there exists $c in mathbb{R}$ such that $|f-c|_{infty} < | f - frac{1}{2}(min f + max f) |_{infty} + varepsilon$



Let $varepsilon > 0$. Define $c =: frac{1}{2}(min f + max f - varepsilon)$.
We have that
$$ |f - c|_{infty} = | f - frac{1}{2}(min f + max f) + frac{1}{2}(min f + max f) -c|_{infty} leq$$
$$ leq | f - frac{1}{2}(min f + max f)|_{infty} + | frac{1}{2}(min f + max f) -c |_{infty}= $$
$$ = | f - frac{1}{2}(min f + max f)|_{infty} + frac{1}{2}varepsilon <$$
$$ < | f - frac{1}{2}(min f + max f)|_{infty} + varepsilon$$






real-analysis functional-analysis proof-verification numerical-methods







share|cite|improve this question















I'm very unsure about what I did here, so if someone could proof check on the following, I would be very thankful.



Let $I$ be a an interval of $mathbb{R}$.



For $f in mathcal{C}(I)$, we say that $P in mathcal{P}_n$ ($P$ is a polynomial of degree $n$) is the best approximation to $f$ if $| f- P |_{infty} = inf , { |f-P|_{infty} : P in mathcal{P}_n } $.



I want to show that the best approximation of $f in mathcal{C}(I)$ by a constant is $frac{1}{2}(inf_{x in I} f(x) + sup_{x in I} f(x))$.



$text{Proof.}$



Since $f$ is continuous and $I$ is compact, $inf f$ and $sup f$ are reached and we have that $inf f < infty$ and $sup f < infty$.



Hence our claim is equivalent to saying that
$inf , { |f - c|_{infty}: c in mathbb{R}} = | f - frac{1}{2}(min f + max f) |_{infty}$



Note that we have that $inf_{c in mathbb{R}} |f - c |_{infty} leq |f - frac{1}{2}(min f + max f) |_{infty}$



Thus it suffices to show that for all $varepsilon > 0$, there exists $c in mathbb{R}$ such that $|f-c|_{infty} < | f - frac{1}{2}(min f + max f) |_{infty} + varepsilon$



Let $varepsilon > 0$. Define $c =: frac{1}{2}(min f + max f - varepsilon)$.
We have that
$$ |f - c|_{infty} = | f - frac{1}{2}(min f + max f) + frac{1}{2}(min f + max f) -c|_{infty} leq$$
$$ leq | f - frac{1}{2}(min f + max f)|_{infty} + | frac{1}{2}(min f + max f) -c |_{infty}= $$
$$ = | f - frac{1}{2}(min f + max f)|_{infty} + frac{1}{2}varepsilon <$$
$$ < | f - frac{1}{2}(min f + max f)|_{infty} + varepsilon$$






real-analysis functional-analysis proof-verification numerical-methods




real-analysis functional-analysis proof-verification numerical-methods






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edited Dec 8 at 22:48

























asked Dec 8 at 22:05









riri92

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  • 1




    Closed and bounded does not imply interval without connectedness. More importantly, from your chain of inequalities, we only get back to $inf |f-c| le | f - ... | $, and the other direction is still needed ( $inf |f-c| ge | f - ... |$)
    – Calvin Khor
    Dec 8 at 22:23










  • could you give me a hint on how to show the other inequality ?
    – riri92
    Dec 8 at 22:29














  • 1




    Closed and bounded does not imply interval without connectedness. More importantly, from your chain of inequalities, we only get back to $inf |f-c| le | f - ... | $, and the other direction is still needed ( $inf |f-c| ge | f - ... |$)
    – Calvin Khor
    Dec 8 at 22:23










  • could you give me a hint on how to show the other inequality ?
    – riri92
    Dec 8 at 22:29








1




1




Closed and bounded does not imply interval without connectedness. More importantly, from your chain of inequalities, we only get back to $inf |f-c| le | f - ... | $, and the other direction is still needed ( $inf |f-c| ge | f - ... |$)
– Calvin Khor
Dec 8 at 22:23




Closed and bounded does not imply interval without connectedness. More importantly, from your chain of inequalities, we only get back to $inf |f-c| le | f - ... | $, and the other direction is still needed ( $inf |f-c| ge | f - ... |$)
– Calvin Khor
Dec 8 at 22:23












could you give me a hint on how to show the other inequality ?
– riri92
Dec 8 at 22:29




could you give me a hint on how to show the other inequality ?
– riri92
Dec 8 at 22:29










2 Answers
2






active

oldest

votes


















3














Hint- By adding a constant to $f$, we may assume that $min f=0$. By multiplying $f$ by a positive constant, we may assume that $max f=1$. You correctly point out that
$ inf_c |f-c| le |f-1/2|.$
The goal now is to prove that $$inf_c |f-c| ge |f-1/2|.$$ Let $c$ be arbitrary. Note that
$|f-c| ge |c|$
and $|f-c| ge |1-c|$. Thus (why?)
$|f-c|ge frac12$. So we are done if we can show
$$ |f-1/2|=1/2.$$
Why is this true?






share|cite|improve this answer























  • Thank you Calvin, I think I can justify all of the above. I still have the following question. Why can't I use the following theorem: Let $S$ be a nonempty subset of the real numbers that is bounded below. The lower bound $w$ is said to be the infimum of $S$ if and only if $forall epsilon>0$ there exists an element $x_{epsilon} in S$ such that $x_{epsilon} <w+ epsilon$. That's what I tried to use in my proof above.
    – riri92
    Dec 8 at 22:53








  • 1




    @riri92 You can, and to apply this result, you need to check that $w$ is a lower bound; this is what I have laid out above.
    – Calvin Khor
    Dec 8 at 23:00








  • 1




    @riri92 yes, that would be enough
    – Calvin Khor
    Dec 8 at 23:05






  • 1




    thank you a lot for your help !
    – riri92
    Dec 8 at 23:06






  • 1




    @riri92 yes, the set is clearly bounded from below. But you need to check that $w$ is a lower bound.
    – Calvin Khor
    Dec 8 at 23:31





















3














Without loss of generality, take the domain of $f$ to be $[0,1]$ and $min_{xin [0,1]} f(x)=0.$ Then, $||f - frac{1}{2}(min f + max f) ||=|(f-frac{1}{2}|f||)|.$ From the definition of the $sup$ norm, noting that $f-frac{1}{2}|f||$ is just a vertical translation of $f$, we have that $|(f-frac{1}{2}|f||)|=|f|-frac{1}{2}|f||=frac{1}{2}|f||$. The result follows.






share|cite|improve this answer





















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    2 Answers
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    2 Answers
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    active

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    3














    Hint- By adding a constant to $f$, we may assume that $min f=0$. By multiplying $f$ by a positive constant, we may assume that $max f=1$. You correctly point out that
    $ inf_c |f-c| le |f-1/2|.$
    The goal now is to prove that $$inf_c |f-c| ge |f-1/2|.$$ Let $c$ be arbitrary. Note that
    $|f-c| ge |c|$
    and $|f-c| ge |1-c|$. Thus (why?)
    $|f-c|ge frac12$. So we are done if we can show
    $$ |f-1/2|=1/2.$$
    Why is this true?






    share|cite|improve this answer























    • Thank you Calvin, I think I can justify all of the above. I still have the following question. Why can't I use the following theorem: Let $S$ be a nonempty subset of the real numbers that is bounded below. The lower bound $w$ is said to be the infimum of $S$ if and only if $forall epsilon>0$ there exists an element $x_{epsilon} in S$ such that $x_{epsilon} <w+ epsilon$. That's what I tried to use in my proof above.
      – riri92
      Dec 8 at 22:53








    • 1




      @riri92 You can, and to apply this result, you need to check that $w$ is a lower bound; this is what I have laid out above.
      – Calvin Khor
      Dec 8 at 23:00








    • 1




      @riri92 yes, that would be enough
      – Calvin Khor
      Dec 8 at 23:05






    • 1




      thank you a lot for your help !
      – riri92
      Dec 8 at 23:06






    • 1




      @riri92 yes, the set is clearly bounded from below. But you need to check that $w$ is a lower bound.
      – Calvin Khor
      Dec 8 at 23:31


















    3














    Hint- By adding a constant to $f$, we may assume that $min f=0$. By multiplying $f$ by a positive constant, we may assume that $max f=1$. You correctly point out that
    $ inf_c |f-c| le |f-1/2|.$
    The goal now is to prove that $$inf_c |f-c| ge |f-1/2|.$$ Let $c$ be arbitrary. Note that
    $|f-c| ge |c|$
    and $|f-c| ge |1-c|$. Thus (why?)
    $|f-c|ge frac12$. So we are done if we can show
    $$ |f-1/2|=1/2.$$
    Why is this true?






    share|cite|improve this answer























    • Thank you Calvin, I think I can justify all of the above. I still have the following question. Why can't I use the following theorem: Let $S$ be a nonempty subset of the real numbers that is bounded below. The lower bound $w$ is said to be the infimum of $S$ if and only if $forall epsilon>0$ there exists an element $x_{epsilon} in S$ such that $x_{epsilon} <w+ epsilon$. That's what I tried to use in my proof above.
      – riri92
      Dec 8 at 22:53








    • 1




      @riri92 You can, and to apply this result, you need to check that $w$ is a lower bound; this is what I have laid out above.
      – Calvin Khor
      Dec 8 at 23:00








    • 1




      @riri92 yes, that would be enough
      – Calvin Khor
      Dec 8 at 23:05






    • 1




      thank you a lot for your help !
      – riri92
      Dec 8 at 23:06






    • 1




      @riri92 yes, the set is clearly bounded from below. But you need to check that $w$ is a lower bound.
      – Calvin Khor
      Dec 8 at 23:31
















    3












    3








    3






    Hint- By adding a constant to $f$, we may assume that $min f=0$. By multiplying $f$ by a positive constant, we may assume that $max f=1$. You correctly point out that
    $ inf_c |f-c| le |f-1/2|.$
    The goal now is to prove that $$inf_c |f-c| ge |f-1/2|.$$ Let $c$ be arbitrary. Note that
    $|f-c| ge |c|$
    and $|f-c| ge |1-c|$. Thus (why?)
    $|f-c|ge frac12$. So we are done if we can show
    $$ |f-1/2|=1/2.$$
    Why is this true?






    share|cite|improve this answer














    Hint- By adding a constant to $f$, we may assume that $min f=0$. By multiplying $f$ by a positive constant, we may assume that $max f=1$. You correctly point out that
    $ inf_c |f-c| le |f-1/2|.$
    The goal now is to prove that $$inf_c |f-c| ge |f-1/2|.$$ Let $c$ be arbitrary. Note that
    $|f-c| ge |c|$
    and $|f-c| ge |1-c|$. Thus (why?)
    $|f-c|ge frac12$. So we are done if we can show
    $$ |f-1/2|=1/2.$$
    Why is this true?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 8 at 22:53

























    answered Dec 8 at 22:48









    Calvin Khor

    11.2k21438




    11.2k21438












    • Thank you Calvin, I think I can justify all of the above. I still have the following question. Why can't I use the following theorem: Let $S$ be a nonempty subset of the real numbers that is bounded below. The lower bound $w$ is said to be the infimum of $S$ if and only if $forall epsilon>0$ there exists an element $x_{epsilon} in S$ such that $x_{epsilon} <w+ epsilon$. That's what I tried to use in my proof above.
      – riri92
      Dec 8 at 22:53








    • 1




      @riri92 You can, and to apply this result, you need to check that $w$ is a lower bound; this is what I have laid out above.
      – Calvin Khor
      Dec 8 at 23:00








    • 1




      @riri92 yes, that would be enough
      – Calvin Khor
      Dec 8 at 23:05






    • 1




      thank you a lot for your help !
      – riri92
      Dec 8 at 23:06






    • 1




      @riri92 yes, the set is clearly bounded from below. But you need to check that $w$ is a lower bound.
      – Calvin Khor
      Dec 8 at 23:31




















    • Thank you Calvin, I think I can justify all of the above. I still have the following question. Why can't I use the following theorem: Let $S$ be a nonempty subset of the real numbers that is bounded below. The lower bound $w$ is said to be the infimum of $S$ if and only if $forall epsilon>0$ there exists an element $x_{epsilon} in S$ such that $x_{epsilon} <w+ epsilon$. That's what I tried to use in my proof above.
      – riri92
      Dec 8 at 22:53








    • 1




      @riri92 You can, and to apply this result, you need to check that $w$ is a lower bound; this is what I have laid out above.
      – Calvin Khor
      Dec 8 at 23:00








    • 1




      @riri92 yes, that would be enough
      – Calvin Khor
      Dec 8 at 23:05






    • 1




      thank you a lot for your help !
      – riri92
      Dec 8 at 23:06






    • 1




      @riri92 yes, the set is clearly bounded from below. But you need to check that $w$ is a lower bound.
      – Calvin Khor
      Dec 8 at 23:31


















    Thank you Calvin, I think I can justify all of the above. I still have the following question. Why can't I use the following theorem: Let $S$ be a nonempty subset of the real numbers that is bounded below. The lower bound $w$ is said to be the infimum of $S$ if and only if $forall epsilon>0$ there exists an element $x_{epsilon} in S$ such that $x_{epsilon} <w+ epsilon$. That's what I tried to use in my proof above.
    – riri92
    Dec 8 at 22:53






    Thank you Calvin, I think I can justify all of the above. I still have the following question. Why can't I use the following theorem: Let $S$ be a nonempty subset of the real numbers that is bounded below. The lower bound $w$ is said to be the infimum of $S$ if and only if $forall epsilon>0$ there exists an element $x_{epsilon} in S$ such that $x_{epsilon} <w+ epsilon$. That's what I tried to use in my proof above.
    – riri92
    Dec 8 at 22:53






    1




    1




    @riri92 You can, and to apply this result, you need to check that $w$ is a lower bound; this is what I have laid out above.
    – Calvin Khor
    Dec 8 at 23:00






    @riri92 You can, and to apply this result, you need to check that $w$ is a lower bound; this is what I have laid out above.
    – Calvin Khor
    Dec 8 at 23:00






    1




    1




    @riri92 yes, that would be enough
    – Calvin Khor
    Dec 8 at 23:05




    @riri92 yes, that would be enough
    – Calvin Khor
    Dec 8 at 23:05




    1




    1




    thank you a lot for your help !
    – riri92
    Dec 8 at 23:06




    thank you a lot for your help !
    – riri92
    Dec 8 at 23:06




    1




    1




    @riri92 yes, the set is clearly bounded from below. But you need to check that $w$ is a lower bound.
    – Calvin Khor
    Dec 8 at 23:31






    @riri92 yes, the set is clearly bounded from below. But you need to check that $w$ is a lower bound.
    – Calvin Khor
    Dec 8 at 23:31













    3














    Without loss of generality, take the domain of $f$ to be $[0,1]$ and $min_{xin [0,1]} f(x)=0.$ Then, $||f - frac{1}{2}(min f + max f) ||=|(f-frac{1}{2}|f||)|.$ From the definition of the $sup$ norm, noting that $f-frac{1}{2}|f||$ is just a vertical translation of $f$, we have that $|(f-frac{1}{2}|f||)|=|f|-frac{1}{2}|f||=frac{1}{2}|f||$. The result follows.






    share|cite|improve this answer


























      3














      Without loss of generality, take the domain of $f$ to be $[0,1]$ and $min_{xin [0,1]} f(x)=0.$ Then, $||f - frac{1}{2}(min f + max f) ||=|(f-frac{1}{2}|f||)|.$ From the definition of the $sup$ norm, noting that $f-frac{1}{2}|f||$ is just a vertical translation of $f$, we have that $|(f-frac{1}{2}|f||)|=|f|-frac{1}{2}|f||=frac{1}{2}|f||$. The result follows.






      share|cite|improve this answer
























        3












        3








        3






        Without loss of generality, take the domain of $f$ to be $[0,1]$ and $min_{xin [0,1]} f(x)=0.$ Then, $||f - frac{1}{2}(min f + max f) ||=|(f-frac{1}{2}|f||)|.$ From the definition of the $sup$ norm, noting that $f-frac{1}{2}|f||$ is just a vertical translation of $f$, we have that $|(f-frac{1}{2}|f||)|=|f|-frac{1}{2}|f||=frac{1}{2}|f||$. The result follows.






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        Without loss of generality, take the domain of $f$ to be $[0,1]$ and $min_{xin [0,1]} f(x)=0.$ Then, $||f - frac{1}{2}(min f + max f) ||=|(f-frac{1}{2}|f||)|.$ From the definition of the $sup$ norm, noting that $f-frac{1}{2}|f||$ is just a vertical translation of $f$, we have that $|(f-frac{1}{2}|f||)|=|f|-frac{1}{2}|f||=frac{1}{2}|f||$. The result follows.







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        answered Dec 8 at 22:56









        Matematleta

        9,9172918




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