Proving that 2 functions are bounded.
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Let $f,g:mathbb R -> mathbb R$. Let $h_1$ = $f-g$ and $h_2$ = $f+g$. Suppose $h_1$ and $h_2$ are bounded. Show that $f$ and $g$ are bounded.
I understand the concept of bounded but would've thought you would need to use limits or a derivative to do a proof. How would you accomplish this without one.
real-analysis
edited 1 hour ago
Henno Brandsma
103k346113
asked 1 hour ago
Pablo Tores
386
add a comment |
up vote
2
down vote
favorite
Let $f,g:mathbb R -> mathbb R$. Let $h_1$ = $f-g$ and $h_2$ = $f+g$. Suppose $h_1$ and $h_2$ are bounded. Show that $f$ and $g$ are bounded.
I understand the concept of bounded but would've thought you would need to use limits or a derivative to do a proof. How would you accomplish this without one.
real-analysis
edited 1 hour ago
Henno Brandsma
103k346113
asked 1 hour ago
Pablo Tores
386
1
Express $f,g $ in the form of $h_2, h_1$.
– xbh
1 hour ago
$|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
– Henno Brandsma
1 hour ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f,g:mathbb R -> mathbb R$. Let $h_1$ = $f-g$ and $h_2$ = $f+g$. Suppose $h_1$ and $h_2$ are bounded. Show that $f$ and $g$ are bounded.
I understand the concept of bounded but would've thought you would need to use limits or a derivative to do a proof. How would you accomplish this without one.
real-analysis
edited 1 hour ago
Henno Brandsma
103k346113
asked 1 hour ago
Pablo Tores
386
Let $f,g:mathbb R -> mathbb R$. Let $h_1$ = $f-g$ and $h_2$ = $f+g$. Suppose $h_1$ and $h_2$ are bounded. Show that $f$ and $g$ are bounded.
I understand the concept of bounded but would've thought you would need to use limits or a derivative to do a proof. How would you accomplish this without one.
real-analysis
real-analysis
edited 1 hour ago
Henno Brandsma
103k346113
asked 1 hour ago
Pablo Tores
386
edited 1 hour ago
Henno Brandsma
103k346113
asked 1 hour ago
Pablo Tores
386
edited 1 hour ago
Henno Brandsma
103k346113
edited 1 hour ago
Henno Brandsma
103k346113
edited 1 hour ago
Henno Brandsma
103k346113
103k346113
asked 1 hour ago
Pablo Tores
386
asked 1 hour ago
Pablo Tores
386
asked 1 hour ago
Pablo Tores
386
386
1
Express $f,g $ in the form of $h_2, h_1$.
– xbh
1 hour ago
$|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
– Henno Brandsma
1 hour ago
add a comment |
1
Express $f,g $ in the form of $h_2, h_1$.
– xbh
1 hour ago
$|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
– Henno Brandsma
1 hour ago
1
1
Express $f,g $ in the form of $h_2, h_1$.
– xbh
1 hour ago
Express $f,g $ in the form of $h_2, h_1$.
– xbh
1 hour ago
$|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
– Henno Brandsma
1 hour ago
$|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
– Henno Brandsma
1 hour ago
add a comment |
2 Answers
2
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up vote
5
down vote
The sum of bounded functions is bounded and so are scalar multiples. Note that $f= frac{h_1 + h_2}{2}$.
Likewise for differences of bounded functions and $g = frac{h_2 - h_1}{2}$.
answered 1 hour ago
Henno Brandsma
103k346113
add a comment |
up vote
1
down vote
Suppose $h_1, h_2$ are bounded. Let $D(f)$ denote the domain of $f$, where $f$ is a function. Then $exists M_1, M_2$ such that $$h_1(x) = f(x) - g(x) leq M_1 forall x in D(h_1), $$ $$h_2(x) = f(x) + g(x) leq M_2 forall x in D(h_2).$$ So, we have that for arbitrary $x in D(f) cap D(g)$, $$ f(x) - g(x) + f(x) + g(x) = 2f(x) leq M_1 + M_2 implies f(x) leq frac{M_1 + M_2}{2}, $$
so $f$ is bounded, by definition. Similarly, notice that
$$ g(x) - f(x) + g(x) + f(x) = 2g(x) leq M_2 - M_1 implies g(x) leq frac{M_2 - M_1}{2}, $$
so $g$ is bounded, by definition. This completes the proof.
answered 54 mins ago
alexsieusahai
636
1
Maybe it should be $xin D(h_1)cap D(h_2)$?
– Shubham Johri
37 mins ago
Ah, you are right; good catch! I've edited my answer to include this.
– alexsieusahai
33 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The sum of bounded functions is bounded and so are scalar multiples. Note that $f= frac{h_1 + h_2}{2}$.
Likewise for differences of bounded functions and $g = frac{h_2 - h_1}{2}$.
answered 1 hour ago
Henno Brandsma
103k346113
add a comment |
up vote
5
down vote
The sum of bounded functions is bounded and so are scalar multiples. Note that $f= frac{h_1 + h_2}{2}$.
Likewise for differences of bounded functions and $g = frac{h_2 - h_1}{2}$.
answered 1 hour ago
Henno Brandsma
103k346113
add a comment |
up vote
5
down vote
up vote
5
down vote
The sum of bounded functions is bounded and so are scalar multiples. Note that $f= frac{h_1 + h_2}{2}$.
Likewise for differences of bounded functions and $g = frac{h_2 - h_1}{2}$.
answered 1 hour ago
Henno Brandsma
103k346113
The sum of bounded functions is bounded and so are scalar multiples. Note that $f= frac{h_1 + h_2}{2}$.
Likewise for differences of bounded functions and $g = frac{h_2 - h_1}{2}$.
answered 1 hour ago
Henno Brandsma
103k346113
answered 1 hour ago
Henno Brandsma
103k346113
answered 1 hour ago
Henno Brandsma
103k346113
answered 1 hour ago
Henno Brandsma
103k346113
103k346113
add a comment |
add a comment |
up vote
1
down vote
Suppose $h_1, h_2$ are bounded. Let $D(f)$ denote the domain of $f$, where $f$ is a function. Then $exists M_1, M_2$ such that $$h_1(x) = f(x) - g(x) leq M_1 forall x in D(h_1), $$ $$h_2(x) = f(x) + g(x) leq M_2 forall x in D(h_2).$$ So, we have that for arbitrary $x in D(f) cap D(g)$, $$ f(x) - g(x) + f(x) + g(x) = 2f(x) leq M_1 + M_2 implies f(x) leq frac{M_1 + M_2}{2}, $$
so $f$ is bounded, by definition. Similarly, notice that
$$ g(x) - f(x) + g(x) + f(x) = 2g(x) leq M_2 - M_1 implies g(x) leq frac{M_2 - M_1}{2}, $$
so $g$ is bounded, by definition. This completes the proof.
answered 54 mins ago
alexsieusahai
636
1
Maybe it should be $xin D(h_1)cap D(h_2)$?
– Shubham Johri
37 mins ago
Ah, you are right; good catch! I've edited my answer to include this.
– alexsieusahai
33 mins ago
add a comment |
up vote
1
down vote
Suppose $h_1, h_2$ are bounded. Let $D(f)$ denote the domain of $f$, where $f$ is a function. Then $exists M_1, M_2$ such that $$h_1(x) = f(x) - g(x) leq M_1 forall x in D(h_1), $$ $$h_2(x) = f(x) + g(x) leq M_2 forall x in D(h_2).$$ So, we have that for arbitrary $x in D(f) cap D(g)$, $$ f(x) - g(x) + f(x) + g(x) = 2f(x) leq M_1 + M_2 implies f(x) leq frac{M_1 + M_2}{2}, $$
so $f$ is bounded, by definition. Similarly, notice that
$$ g(x) - f(x) + g(x) + f(x) = 2g(x) leq M_2 - M_1 implies g(x) leq frac{M_2 - M_1}{2}, $$
so $g$ is bounded, by definition. This completes the proof.
answered 54 mins ago
alexsieusahai
636
1
Maybe it should be $xin D(h_1)cap D(h_2)$?
– Shubham Johri
37 mins ago
Ah, you are right; good catch! I've edited my answer to include this.
– alexsieusahai
33 mins ago
add a comment |
up vote
1
down vote
up vote
1
down vote
Suppose $h_1, h_2$ are bounded. Let $D(f)$ denote the domain of $f$, where $f$ is a function. Then $exists M_1, M_2$ such that $$h_1(x) = f(x) - g(x) leq M_1 forall x in D(h_1), $$ $$h_2(x) = f(x) + g(x) leq M_2 forall x in D(h_2).$$ So, we have that for arbitrary $x in D(f) cap D(g)$, $$ f(x) - g(x) + f(x) + g(x) = 2f(x) leq M_1 + M_2 implies f(x) leq frac{M_1 + M_2}{2}, $$
so $f$ is bounded, by definition. Similarly, notice that
$$ g(x) - f(x) + g(x) + f(x) = 2g(x) leq M_2 - M_1 implies g(x) leq frac{M_2 - M_1}{2}, $$
so $g$ is bounded, by definition. This completes the proof.
answered 54 mins ago
alexsieusahai
636
Suppose $h_1, h_2$ are bounded. Let $D(f)$ denote the domain of $f$, where $f$ is a function. Then $exists M_1, M_2$ such that $$h_1(x) = f(x) - g(x) leq M_1 forall x in D(h_1), $$ $$h_2(x) = f(x) + g(x) leq M_2 forall x in D(h_2).$$ So, we have that for arbitrary $x in D(f) cap D(g)$, $$ f(x) - g(x) + f(x) + g(x) = 2f(x) leq M_1 + M_2 implies f(x) leq frac{M_1 + M_2}{2}, $$
so $f$ is bounded, by definition. Similarly, notice that
$$ g(x) - f(x) + g(x) + f(x) = 2g(x) leq M_2 - M_1 implies g(x) leq frac{M_2 - M_1}{2}, $$
so $g$ is bounded, by definition. This completes the proof.
answered 54 mins ago
alexsieusahai
636
edited 33 mins ago
answered 54 mins ago
alexsieusahai
636
answered 54 mins ago
alexsieusahai
636
answered 54 mins ago
alexsieusahai
636
636
1
Maybe it should be $xin D(h_1)cap D(h_2)$?
– Shubham Johri
37 mins ago
Ah, you are right; good catch! I've edited my answer to include this.
– alexsieusahai
33 mins ago
add a comment |
1
Maybe it should be $xin D(h_1)cap D(h_2)$?
– Shubham Johri
37 mins ago
Ah, you are right; good catch! I've edited my answer to include this.
– alexsieusahai
33 mins ago
1
1
Maybe it should be $xin D(h_1)cap D(h_2)$?
– Shubham Johri
37 mins ago
Maybe it should be $xin D(h_1)cap D(h_2)$?
– Shubham Johri
37 mins ago
Ah, you are right; good catch! I've edited my answer to include this.
– alexsieusahai
33 mins ago
Ah, you are right; good catch! I've edited my answer to include this.
– alexsieusahai
33 mins ago
add a comment |
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1
Express $f,g $ in the form of $h_2, h_1$.
– xbh
1 hour ago
$|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
– Henno Brandsma
1 hour ago