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Prove a function f is uniformly bounded throughout [0, ∞)
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Let f : [0, ∞) → [0, ∞) be a continuous function with f(0) = 0. Show that if $f(t) ≤ 1 +f(t)^3/10 $ for all t ∈ [0, ∞), then f is uniformly bounded throughout [0, ∞)
I am wondering if the function is uniformly bounded since the definition of uniformly bounded is for a set of functions. And Suppose f(t)= any positive large number (like greater than 10), it seems to be always true that $f(t) ≤ 1 + f(t)^3/10 $.
I am also considering if I can prove the function is uniformly continuous to show it is uniformly bounded.
real-analysis
asked Jan 25 at 17:42
Eric JEric J
163
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add a comment |
$begingroup$
Let f : [0, ∞) → [0, ∞) be a continuous function with f(0) = 0. Show that if $f(t) ≤ 1 +f(t)^3/10 $ for all t ∈ [0, ∞), then f is uniformly bounded throughout [0, ∞)
I am wondering if the function is uniformly bounded since the definition of uniformly bounded is for a set of functions. And Suppose f(t)= any positive large number (like greater than 10), it seems to be always true that $f(t) ≤ 1 + f(t)^3/10 $.
I am also considering if I can prove the function is uniformly continuous to show it is uniformly bounded.
real-analysis
asked Jan 25 at 17:42
Eric JEric J
163
$endgroup$
add a comment |
$begingroup$
Let f : [0, ∞) → [0, ∞) be a continuous function with f(0) = 0. Show that if $f(t) ≤ 1 +f(t)^3/10 $ for all t ∈ [0, ∞), then f is uniformly bounded throughout [0, ∞)
I am wondering if the function is uniformly bounded since the definition of uniformly bounded is for a set of functions. And Suppose f(t)= any positive large number (like greater than 10), it seems to be always true that $f(t) ≤ 1 + f(t)^3/10 $.
I am also considering if I can prove the function is uniformly continuous to show it is uniformly bounded.
real-analysis
asked Jan 25 at 17:42
Eric JEric J
163
$endgroup$
Let f : [0, ∞) → [0, ∞) be a continuous function with f(0) = 0. Show that if $f(t) ≤ 1 +f(t)^3/10 $ for all t ∈ [0, ∞), then f is uniformly bounded throughout [0, ∞)
I am wondering if the function is uniformly bounded since the definition of uniformly bounded is for a set of functions. And Suppose f(t)= any positive large number (like greater than 10), it seems to be always true that $f(t) ≤ 1 + f(t)^3/10 $.
I am also considering if I can prove the function is uniformly continuous to show it is uniformly bounded.
real-analysis
real-analysis
asked Jan 25 at 17:42
Eric JEric J
163
asked Jan 25 at 17:42
Eric JEric J
163
asked Jan 25 at 17:42
Eric JEric J
163
asked Jan 25 at 17:42
Eric JEric J
163
asked Jan 25 at 17:42
Eric JEric J
163
163
add a comment |
add a comment |
1 Answer
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$begingroup$
Note that $2 > 1+2^3/10$, which entails that $f$ is never equal to $2$. By IVT, $f$ remains in $[0,2)$ so is bounded.
answered Jan 25 at 17:46
MindlackMindlack
4,920211
$endgroup$
$begingroup$
It is f(t)≤1+f(t)^3/10, hence f(x) cannot be 2
$endgroup$
– Eric J
Jan 25 at 17:50
1
$begingroup$
That is exactly what I wrote.
$endgroup$
– Mindlack
Jan 25 at 17:53
add a comment |
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1 Answer
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$begingroup$
Note that $2 > 1+2^3/10$, which entails that $f$ is never equal to $2$. By IVT, $f$ remains in $[0,2)$ so is bounded.
answered Jan 25 at 17:46
MindlackMindlack
4,920211
$endgroup$
$begingroup$
It is f(t)≤1+f(t)^3/10, hence f(x) cannot be 2
$endgroup$
– Eric J
Jan 25 at 17:50
1
$begingroup$
That is exactly what I wrote.
$endgroup$
– Mindlack
Jan 25 at 17:53
add a comment |
$begingroup$
Note that $2 > 1+2^3/10$, which entails that $f$ is never equal to $2$. By IVT, $f$ remains in $[0,2)$ so is bounded.
answered Jan 25 at 17:46
MindlackMindlack
4,920211
$endgroup$
$begingroup$
It is f(t)≤1+f(t)^3/10, hence f(x) cannot be 2
$endgroup$
– Eric J
Jan 25 at 17:50
1
$begingroup$
That is exactly what I wrote.
$endgroup$
– Mindlack
Jan 25 at 17:53
add a comment |
$begingroup$
Note that $2 > 1+2^3/10$, which entails that $f$ is never equal to $2$. By IVT, $f$ remains in $[0,2)$ so is bounded.
answered Jan 25 at 17:46
MindlackMindlack
4,920211
$endgroup$
Note that $2 > 1+2^3/10$, which entails that $f$ is never equal to $2$. By IVT, $f$ remains in $[0,2)$ so is bounded.
answered Jan 25 at 17:46
MindlackMindlack
4,920211
answered Jan 25 at 17:46
MindlackMindlack
4,920211
answered Jan 25 at 17:46
MindlackMindlack
4,920211
answered Jan 25 at 17:46
MindlackMindlack
4,920211
4,920211
$begingroup$
It is f(t)≤1+f(t)^3/10, hence f(x) cannot be 2
$endgroup$
– Eric J
Jan 25 at 17:50
1
$begingroup$
That is exactly what I wrote.
$endgroup$
– Mindlack
Jan 25 at 17:53
add a comment |
$begingroup$
It is f(t)≤1+f(t)^3/10, hence f(x) cannot be 2
$endgroup$
– Eric J
Jan 25 at 17:50
1
$begingroup$
That is exactly what I wrote.
$endgroup$
– Mindlack
Jan 25 at 17:53
$begingroup$
It is f(t)≤1+f(t)^3/10, hence f(x) cannot be 2
$endgroup$
– Eric J
Jan 25 at 17:50
$begingroup$
It is f(t)≤1+f(t)^3/10, hence f(x) cannot be 2
$endgroup$
– Eric J
Jan 25 at 17:50
1
1
$begingroup$
That is exactly what I wrote.
$endgroup$
– Mindlack
Jan 25 at 17:53
$begingroup$
That is exactly what I wrote.
$endgroup$
– Mindlack
Jan 25 at 17:53
add a comment |
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