How do I show that the sup norm metric is a metric for $Bbb{R}^n$? (the triangular inequality specifically)
$begingroup$
I am not sure how to approach such inequality. I have shown the other 3 requirement for the sup norm to be a metric for $Bbb{R}^n$.
For $x,y,zin Bbb{R}^n$, we need to show that
$$sup{lvert x_i - z_irvert }leq sup{lvert x_i - y_i rvert } + sup{lvert y_i - z_i rvert }text{ for }iin {1,2,...,n}$$
My thought is that this should be related to the cauchy schwarz inequality. But not sure how to start. since C.S is in the series form not in the supremum form.
Thanks for help
analysis
edited Jan 15 at 2:33
Theo Bendit
20.1k12354
asked Jan 15 at 2:23
RicoRico
908
$endgroup$
add a comment |
$begingroup$
I am not sure how to approach such inequality. I have shown the other 3 requirement for the sup norm to be a metric for $Bbb{R}^n$.
For $x,y,zin Bbb{R}^n$, we need to show that
$$sup{lvert x_i - z_irvert }leq sup{lvert x_i - y_i rvert } + sup{lvert y_i - z_i rvert }text{ for }iin {1,2,...,n}$$
My thought is that this should be related to the cauchy schwarz inequality. But not sure how to start. since C.S is in the series form not in the supremum form.
Thanks for help
analysis
edited Jan 15 at 2:33
Theo Bendit
20.1k12354
asked Jan 15 at 2:23
RicoRico
908
$endgroup$
$begingroup$
Hint: note that $$sup_i (|x_i - y_i| + |y_i - z_i|) le sup_{i,j} (|x_i - y_i| + |y_j - z_j|) = sup_i |x_i - y_i| + sup_j |y_j - z_j|.$$
$endgroup$
– Theo Bendit
Jan 15 at 2:28
$begingroup$
Let me know if you'd like a full answer.
$endgroup$
– Theo Bendit
Jan 15 at 2:40
$begingroup$
May I ask why the absolute value is not necessary? I am following the textbook definition where it has the absolute value in there, also since we are considering $R^n$ shouldn't there be negative numbers? Sorry this is probably a trivial question but I am trying to learn. Thank you so much for helping me
$endgroup$
– Rico
Jan 15 at 2:40
$begingroup$
What I meant in my deleted comment is that, in order to show the above inequality is true, you shouldn't need any properties specific to absolute values, as this holds for more general functions. If you are computing the supremum of the sum of two functions, this will generally be less than if you compute the suprema of the functions separately and add the result. The condition that $x = y$ when maximising $f(x) + g(y)$ is a restriction, and hence yields a smaller result than when maximising $f(x) + g(y)$ with $x$ and $y$ unrestricted.
$endgroup$
– Theo Bendit
Jan 15 at 2:43
add a comment |
$begingroup$
I am not sure how to approach such inequality. I have shown the other 3 requirement for the sup norm to be a metric for $Bbb{R}^n$.
For $x,y,zin Bbb{R}^n$, we need to show that
$$sup{lvert x_i - z_irvert }leq sup{lvert x_i - y_i rvert } + sup{lvert y_i - z_i rvert }text{ for }iin {1,2,...,n}$$
My thought is that this should be related to the cauchy schwarz inequality. But not sure how to start. since C.S is in the series form not in the supremum form.
Thanks for help
analysis
edited Jan 15 at 2:33
Theo Bendit
20.1k12354
asked Jan 15 at 2:23
RicoRico
908
$endgroup$
I am not sure how to approach such inequality. I have shown the other 3 requirement for the sup norm to be a metric for $Bbb{R}^n$.
For $x,y,zin Bbb{R}^n$, we need to show that
$$sup{lvert x_i - z_irvert }leq sup{lvert x_i - y_i rvert } + sup{lvert y_i - z_i rvert }text{ for }iin {1,2,...,n}$$
My thought is that this should be related to the cauchy schwarz inequality. But not sure how to start. since C.S is in the series form not in the supremum form.
Thanks for help
analysis
analysis
edited Jan 15 at 2:33
Theo Bendit
20.1k12354
asked Jan 15 at 2:23
RicoRico
908
edited Jan 15 at 2:33
Theo Bendit
20.1k12354
asked Jan 15 at 2:23
RicoRico
908
edited Jan 15 at 2:33
Theo Bendit
20.1k12354
edited Jan 15 at 2:33
Theo Bendit
20.1k12354
edited Jan 15 at 2:33
Theo Bendit
20.1k12354
20.1k12354
asked Jan 15 at 2:23
RicoRico
908
asked Jan 15 at 2:23
RicoRico
908
asked Jan 15 at 2:23
RicoRico
908
908
$begingroup$
Hint: note that $$sup_i (|x_i - y_i| + |y_i - z_i|) le sup_{i,j} (|x_i - y_i| + |y_j - z_j|) = sup_i |x_i - y_i| + sup_j |y_j - z_j|.$$
$endgroup$
– Theo Bendit
Jan 15 at 2:28
$begingroup$
Let me know if you'd like a full answer.
$endgroup$
– Theo Bendit
Jan 15 at 2:40
$begingroup$
May I ask why the absolute value is not necessary? I am following the textbook definition where it has the absolute value in there, also since we are considering $R^n$ shouldn't there be negative numbers? Sorry this is probably a trivial question but I am trying to learn. Thank you so much for helping me
$endgroup$
– Rico
Jan 15 at 2:40
$begingroup$
What I meant in my deleted comment is that, in order to show the above inequality is true, you shouldn't need any properties specific to absolute values, as this holds for more general functions. If you are computing the supremum of the sum of two functions, this will generally be less than if you compute the suprema of the functions separately and add the result. The condition that $x = y$ when maximising $f(x) + g(y)$ is a restriction, and hence yields a smaller result than when maximising $f(x) + g(y)$ with $x$ and $y$ unrestricted.
$endgroup$
– Theo Bendit
Jan 15 at 2:43
add a comment |
$begingroup$
Hint: note that $$sup_i (|x_i - y_i| + |y_i - z_i|) le sup_{i,j} (|x_i - y_i| + |y_j - z_j|) = sup_i |x_i - y_i| + sup_j |y_j - z_j|.$$
$endgroup$
– Theo Bendit
Jan 15 at 2:28
$begingroup$
Let me know if you'd like a full answer.
$endgroup$
– Theo Bendit
Jan 15 at 2:40
$begingroup$
May I ask why the absolute value is not necessary? I am following the textbook definition where it has the absolute value in there, also since we are considering $R^n$ shouldn't there be negative numbers? Sorry this is probably a trivial question but I am trying to learn. Thank you so much for helping me
$endgroup$
– Rico
Jan 15 at 2:40
$begingroup$
What I meant in my deleted comment is that, in order to show the above inequality is true, you shouldn't need any properties specific to absolute values, as this holds for more general functions. If you are computing the supremum of the sum of two functions, this will generally be less than if you compute the suprema of the functions separately and add the result. The condition that $x = y$ when maximising $f(x) + g(y)$ is a restriction, and hence yields a smaller result than when maximising $f(x) + g(y)$ with $x$ and $y$ unrestricted.
$endgroup$
– Theo Bendit
Jan 15 at 2:43
$begingroup$
Hint: note that $$sup_i (|x_i - y_i| + |y_i - z_i|) le sup_{i,j} (|x_i - y_i| + |y_j - z_j|) = sup_i |x_i - y_i| + sup_j |y_j - z_j|.$$
$endgroup$
– Theo Bendit
Jan 15 at 2:28
$begingroup$
Hint: note that $$sup_i (|x_i - y_i| + |y_i - z_i|) le sup_{i,j} (|x_i - y_i| + |y_j - z_j|) = sup_i |x_i - y_i| + sup_j |y_j - z_j|.$$
$endgroup$
– Theo Bendit
Jan 15 at 2:28
$begingroup$
Let me know if you'd like a full answer.
$endgroup$
– Theo Bendit
Jan 15 at 2:40
$begingroup$
Let me know if you'd like a full answer.
$endgroup$
– Theo Bendit
Jan 15 at 2:40
$begingroup$
May I ask why the absolute value is not necessary? I am following the textbook definition where it has the absolute value in there, also since we are considering $R^n$ shouldn't there be negative numbers? Sorry this is probably a trivial question but I am trying to learn. Thank you so much for helping me
$endgroup$
– Rico
Jan 15 at 2:40
$begingroup$
May I ask why the absolute value is not necessary? I am following the textbook definition where it has the absolute value in there, also since we are considering $R^n$ shouldn't there be negative numbers? Sorry this is probably a trivial question but I am trying to learn. Thank you so much for helping me
$endgroup$
– Rico
Jan 15 at 2:40
$begingroup$
What I meant in my deleted comment is that, in order to show the above inequality is true, you shouldn't need any properties specific to absolute values, as this holds for more general functions. If you are computing the supremum of the sum of two functions, this will generally be less than if you compute the suprema of the functions separately and add the result. The condition that $x = y$ when maximising $f(x) + g(y)$ is a restriction, and hence yields a smaller result than when maximising $f(x) + g(y)$ with $x$ and $y$ unrestricted.
$endgroup$
– Theo Bendit
Jan 15 at 2:43
$begingroup$
What I meant in my deleted comment is that, in order to show the above inequality is true, you shouldn't need any properties specific to absolute values, as this holds for more general functions. If you are computing the supremum of the sum of two functions, this will generally be less than if you compute the suprema of the functions separately and add the result. The condition that $x = y$ when maximising $f(x) + g(y)$ is a restriction, and hence yields a smaller result than when maximising $f(x) + g(y)$ with $x$ and $y$ unrestricted.
$endgroup$
– Theo Bendit
Jan 15 at 2:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Lemma: Suppose $S$ is a set and $f, g : S to Bbb{R}$. Then
$$sup_{s in S} (f(s) + g(s)) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
Proof: Suppose $S, f, g$ are as in the statement. Note that, for any $s_0 in S$, we have
begin{align*}
f(s_0) le sup_{s in S} f(s) \
g(s_0) le sup_{t in S} g(t),
end{align*}
hence
$$f(s_0) + g(s_0) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
That is, $sup_{s in S} f(s) + sup_{t in S} g(t)$ is an upper bound for ${f(s) + g(s) : s in S}$, hence
$$sup_{s in S} (f(s) + g(s)) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
In particular, apply the lemma with $S = {1, ldots, n}$, $f : i mapsto |x_i - y_i|$, and $g : j mapsto |y_j - z_j|$. Then,
$$sup_i (|x_i - y_i| + |y_i - z_i|) le sup_i |x_i - y_i| + sup_j |y_j - z_j|.$$
Note that, by triangle inequality, we have $|x_i - z_i| le |x_i - y_i| + |y_i - z_i|$, and hence taking the supremum of both sides, we get
$$sup_i |x_i - z_i| le sup_i (|x_i - y_i| + |y_i - z_i|) le sup_i |x_i - y_i| + sup_j |y_j - z_j|,$$
which is what we needed!
answered Jan 15 at 2:54
Theo BenditTheo Bendit
20.1k12354
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– Rico
Jan 15 at 3:31
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Lemma: Suppose $S$ is a set and $f, g : S to Bbb{R}$. Then
$$sup_{s in S} (f(s) + g(s)) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
Proof: Suppose $S, f, g$ are as in the statement. Note that, for any $s_0 in S$, we have
begin{align*}
f(s_0) le sup_{s in S} f(s) \
g(s_0) le sup_{t in S} g(t),
end{align*}
hence
$$f(s_0) + g(s_0) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
That is, $sup_{s in S} f(s) + sup_{t in S} g(t)$ is an upper bound for ${f(s) + g(s) : s in S}$, hence
$$sup_{s in S} (f(s) + g(s)) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
In particular, apply the lemma with $S = {1, ldots, n}$, $f : i mapsto |x_i - y_i|$, and $g : j mapsto |y_j - z_j|$. Then,
$$sup_i (|x_i - y_i| + |y_i - z_i|) le sup_i |x_i - y_i| + sup_j |y_j - z_j|.$$
Note that, by triangle inequality, we have $|x_i - z_i| le |x_i - y_i| + |y_i - z_i|$, and hence taking the supremum of both sides, we get
$$sup_i |x_i - z_i| le sup_i (|x_i - y_i| + |y_i - z_i|) le sup_i |x_i - y_i| + sup_j |y_j - z_j|,$$
which is what we needed!
answered Jan 15 at 2:54
Theo BenditTheo Bendit
20.1k12354
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– Rico
Jan 15 at 3:31
add a comment |
$begingroup$
Lemma: Suppose $S$ is a set and $f, g : S to Bbb{R}$. Then
$$sup_{s in S} (f(s) + g(s)) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
Proof: Suppose $S, f, g$ are as in the statement. Note that, for any $s_0 in S$, we have
begin{align*}
f(s_0) le sup_{s in S} f(s) \
g(s_0) le sup_{t in S} g(t),
end{align*}
hence
$$f(s_0) + g(s_0) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
That is, $sup_{s in S} f(s) + sup_{t in S} g(t)$ is an upper bound for ${f(s) + g(s) : s in S}$, hence
$$sup_{s in S} (f(s) + g(s)) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
In particular, apply the lemma with $S = {1, ldots, n}$, $f : i mapsto |x_i - y_i|$, and $g : j mapsto |y_j - z_j|$. Then,
$$sup_i (|x_i - y_i| + |y_i - z_i|) le sup_i |x_i - y_i| + sup_j |y_j - z_j|.$$
Note that, by triangle inequality, we have $|x_i - z_i| le |x_i - y_i| + |y_i - z_i|$, and hence taking the supremum of both sides, we get
$$sup_i |x_i - z_i| le sup_i (|x_i - y_i| + |y_i - z_i|) le sup_i |x_i - y_i| + sup_j |y_j - z_j|,$$
which is what we needed!
answered Jan 15 at 2:54
Theo BenditTheo Bendit
20.1k12354
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– Rico
Jan 15 at 3:31
add a comment |
$begingroup$
Lemma: Suppose $S$ is a set and $f, g : S to Bbb{R}$. Then
$$sup_{s in S} (f(s) + g(s)) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
Proof: Suppose $S, f, g$ are as in the statement. Note that, for any $s_0 in S$, we have
begin{align*}
f(s_0) le sup_{s in S} f(s) \
g(s_0) le sup_{t in S} g(t),
end{align*}
hence
$$f(s_0) + g(s_0) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
That is, $sup_{s in S} f(s) + sup_{t in S} g(t)$ is an upper bound for ${f(s) + g(s) : s in S}$, hence
$$sup_{s in S} (f(s) + g(s)) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
In particular, apply the lemma with $S = {1, ldots, n}$, $f : i mapsto |x_i - y_i|$, and $g : j mapsto |y_j - z_j|$. Then,
$$sup_i (|x_i - y_i| + |y_i - z_i|) le sup_i |x_i - y_i| + sup_j |y_j - z_j|.$$
Note that, by triangle inequality, we have $|x_i - z_i| le |x_i - y_i| + |y_i - z_i|$, and hence taking the supremum of both sides, we get
$$sup_i |x_i - z_i| le sup_i (|x_i - y_i| + |y_i - z_i|) le sup_i |x_i - y_i| + sup_j |y_j - z_j|,$$
which is what we needed!
answered Jan 15 at 2:54
Theo BenditTheo Bendit
20.1k12354
$endgroup$
Lemma: Suppose $S$ is a set and $f, g : S to Bbb{R}$. Then
$$sup_{s in S} (f(s) + g(s)) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
Proof: Suppose $S, f, g$ are as in the statement. Note that, for any $s_0 in S$, we have
begin{align*}
f(s_0) le sup_{s in S} f(s) \
g(s_0) le sup_{t in S} g(t),
end{align*}
hence
$$f(s_0) + g(s_0) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
That is, $sup_{s in S} f(s) + sup_{t in S} g(t)$ is an upper bound for ${f(s) + g(s) : s in S}$, hence
$$sup_{s in S} (f(s) + g(s)) le sup_{s in S} f(s) + sup_{t in S} g(t).$$
In particular, apply the lemma with $S = {1, ldots, n}$, $f : i mapsto |x_i - y_i|$, and $g : j mapsto |y_j - z_j|$. Then,
$$sup_i (|x_i - y_i| + |y_i - z_i|) le sup_i |x_i - y_i| + sup_j |y_j - z_j|.$$
Note that, by triangle inequality, we have $|x_i - z_i| le |x_i - y_i| + |y_i - z_i|$, and hence taking the supremum of both sides, we get
$$sup_i |x_i - z_i| le sup_i (|x_i - y_i| + |y_i - z_i|) le sup_i |x_i - y_i| + sup_j |y_j - z_j|,$$
which is what we needed!
answered Jan 15 at 2:54
Theo BenditTheo Bendit
20.1k12354
answered Jan 15 at 2:54
Theo BenditTheo Bendit
20.1k12354
answered Jan 15 at 2:54
Theo BenditTheo Bendit
20.1k12354
answered Jan 15 at 2:54
Theo BenditTheo Bendit
20.1k12354
20.1k12354
$begingroup$
Thank you so much!
$endgroup$
– Rico
Jan 15 at 3:31
add a comment |
$begingroup$
Thank you so much!
$endgroup$
– Rico
Jan 15 at 3:31
$begingroup$
Thank you so much!
$endgroup$
– Rico
Jan 15 at 3:31
$begingroup$
Thank you so much!
$endgroup$
– Rico
Jan 15 at 3:31
add a comment |
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$begingroup$
Hint: note that $$sup_i (|x_i - y_i| + |y_i - z_i|) le sup_{i,j} (|x_i - y_i| + |y_j - z_j|) = sup_i |x_i - y_i| + sup_j |y_j - z_j|.$$
$endgroup$
– Theo Bendit
Jan 15 at 2:28
$begingroup$
Let me know if you'd like a full answer.
$endgroup$
– Theo Bendit
Jan 15 at 2:40
$begingroup$
May I ask why the absolute value is not necessary? I am following the textbook definition where it has the absolute value in there, also since we are considering $R^n$ shouldn't there be negative numbers? Sorry this is probably a trivial question but I am trying to learn. Thank you so much for helping me
$endgroup$
– Rico
Jan 15 at 2:40
$begingroup$
What I meant in my deleted comment is that, in order to show the above inequality is true, you shouldn't need any properties specific to absolute values, as this holds for more general functions. If you are computing the supremum of the sum of two functions, this will generally be less than if you compute the suprema of the functions separately and add the result. The condition that $x = y$ when maximising $f(x) + g(y)$ is a restriction, and hence yields a smaller result than when maximising $f(x) + g(y)$ with $x$ and $y$ unrestricted.
$endgroup$
– Theo Bendit
Jan 15 at 2:43