Show that $ inf_{c in mathbb{R}} |f - c|_{infty} = | f - frac{1}{2}(min f + max f) |_{infty}$
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
asked Dec 8 at 22:05
riri92
1657
add a comment |
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
asked Dec 8 at 22:05
riri92
1657
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
add a comment |
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
asked Dec 8 at 22:05
riri92
1657
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
asked Dec 8 at 22:05
riri92
1657
asked Dec 8 at 22:05
riri92
1657
edited Dec 8 at 22:48
asked Dec 8 at 22:05
riri92
1657
asked Dec 8 at 22:05
riri92
1657
asked Dec 8 at 22:05
riri92
1657
1657
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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?
answered Dec 8 at 22:48
Calvin Khor
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
|
show 2 more comments
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.
answered Dec 8 at 22:56
Matematleta
9,9172918
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
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?
answered Dec 8 at 22:48
Calvin Khor
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
|
show 2 more comments
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?
answered Dec 8 at 22:48
Calvin Khor
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
|
show 2 more comments
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?
answered Dec 8 at 22:48
Calvin Khor
11.2k21438
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?
answered Dec 8 at 22:48
Calvin Khor
11.2k21438
edited Dec 8 at 22:53
answered Dec 8 at 22:48
Calvin Khor
11.2k21438
answered Dec 8 at 22:48
Calvin Khor
11.2k21438
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
|
show 2 more comments
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
|
show 2 more comments
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.
answered Dec 8 at 22:56
Matematleta
9,9172918
add a comment |
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.
answered Dec 8 at 22:56
Matematleta
9,9172918
add a comment |
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.
answered Dec 8 at 22:56
Matematleta
9,9172918
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.
answered Dec 8 at 22:56
Matematleta
9,9172918
answered Dec 8 at 22:56
Matematleta
9,9172918
answered Dec 8 at 22:56
Matematleta
9,9172918
answered Dec 8 at 22:56
Matematleta
9,9172918
9,9172918
add a comment |
add a comment |
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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