Minimum value of a continuous function.
Let $f:[a, b]to (0, infty)$ be a continuous function. Let $$F:[a, b]times [a, b]to (0, infty): F(x, y) =frac{f(x)}{f(y)}$$
Then I am interested in the lower bound on $F$.
If it is $1$ then how to show it?
functions continuity upper-lower-bounds
asked Nov 24 at 8:42
Mittal G
1,188515
add a comment |
Let $f:[a, b]to (0, infty)$ be a continuous function. Let $$F:[a, b]times [a, b]to (0, infty): F(x, y) =frac{f(x)}{f(y)}$$
Then I am interested in the lower bound on $F$.
If it is $1$ then how to show it?
functions continuity upper-lower-bounds
asked Nov 24 at 8:42
Mittal G
1,188515
No i think we have to make use of compactness of $[a, b]$.
– Mittal G
Nov 24 at 8:50
add a comment |
Let $f:[a, b]to (0, infty)$ be a continuous function. Let $$F:[a, b]times [a, b]to (0, infty): F(x, y) =frac{f(x)}{f(y)}$$
Then I am interested in the lower bound on $F$.
If it is $1$ then how to show it?
functions continuity upper-lower-bounds
asked Nov 24 at 8:42
Mittal G
1,188515
Let $f:[a, b]to (0, infty)$ be a continuous function. Let $$F:[a, b]times [a, b]to (0, infty): F(x, y) =frac{f(x)}{f(y)}$$
Then I am interested in the lower bound on $F$.
If it is $1$ then how to show it?
functions continuity upper-lower-bounds
functions continuity upper-lower-bounds
asked Nov 24 at 8:42
Mittal G
1,188515
asked Nov 24 at 8:42
Mittal G
1,188515
asked Nov 24 at 8:42
Mittal G
1,188515
asked Nov 24 at 8:42
Mittal G
1,188515
asked Nov 24 at 8:42
Mittal G
1,188515
1,188515
No i think we have to make use of compactness of $[a, b]$.
– Mittal G
Nov 24 at 8:50
add a comment |
No i think we have to make use of compactness of $[a, b]$.
– Mittal G
Nov 24 at 8:50
No i think we have to make use of compactness of $[a, b]$.
– Mittal G
Nov 24 at 8:50
No i think we have to make use of compactness of $[a, b]$.
– Mittal G
Nov 24 at 8:50
add a comment |
1 Answer
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I think we can have a better lower bound that $0$. Since $left[ a, b right]$ is compact and $f$ is continuous, $f$ attains it maximum and minimum value in some interval. Let us name it $M$ and $m$ respectively. Then,
$$m leq f left( x right) leq M$$
This gives $F left( x, y right) geq dfrac{m}{M}$
Hence, $dfrac{m}{M}$ will be your required lower bound.
answered Nov 24 at 8:51
Aniruddha Deshmukh
889418
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think we can have a better lower bound that $0$. Since $left[ a, b right]$ is compact and $f$ is continuous, $f$ attains it maximum and minimum value in some interval. Let us name it $M$ and $m$ respectively. Then,
$$m leq f left( x right) leq M$$
This gives $F left( x, y right) geq dfrac{m}{M}$
Hence, $dfrac{m}{M}$ will be your required lower bound.
answered Nov 24 at 8:51
Aniruddha Deshmukh
889418
add a comment |
I think we can have a better lower bound that $0$. Since $left[ a, b right]$ is compact and $f$ is continuous, $f$ attains it maximum and minimum value in some interval. Let us name it $M$ and $m$ respectively. Then,
$$m leq f left( x right) leq M$$
This gives $F left( x, y right) geq dfrac{m}{M}$
Hence, $dfrac{m}{M}$ will be your required lower bound.
answered Nov 24 at 8:51
Aniruddha Deshmukh
889418
add a comment |
I think we can have a better lower bound that $0$. Since $left[ a, b right]$ is compact and $f$ is continuous, $f$ attains it maximum and minimum value in some interval. Let us name it $M$ and $m$ respectively. Then,
$$m leq f left( x right) leq M$$
This gives $F left( x, y right) geq dfrac{m}{M}$
Hence, $dfrac{m}{M}$ will be your required lower bound.
answered Nov 24 at 8:51
Aniruddha Deshmukh
889418
I think we can have a better lower bound that $0$. Since $left[ a, b right]$ is compact and $f$ is continuous, $f$ attains it maximum and minimum value in some interval. Let us name it $M$ and $m$ respectively. Then,
$$m leq f left( x right) leq M$$
This gives $F left( x, y right) geq dfrac{m}{M}$
Hence, $dfrac{m}{M}$ will be your required lower bound.
answered Nov 24 at 8:51
Aniruddha Deshmukh
889418
answered Nov 24 at 8:51
Aniruddha Deshmukh
889418
answered Nov 24 at 8:51
Aniruddha Deshmukh
889418
answered Nov 24 at 8:51
Aniruddha Deshmukh
889418
889418
add a comment |
add a comment |
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No i think we have to make use of compactness of $[a, b]$.
– Mittal G
Nov 24 at 8:50