Does a bounded function $f$ on $mathbb R$ has zero for its second derivative?
Ask: Does a bounded function $f$ on $mathbb R$, which is continuous and twice differentiable on $mathbb R$, has zero for its second derivative, i.e. there exists a point $x_0 in mathbb R$ such that
$$f''(x_0)=0$$
My idea is considering the contrapositive for the statement above. Let say
$$f''(x_0)>0$$
, or equivalently the function $f$ is strictly convex. (Another case: Consider the function $-f$, which is strictly concave and having $-f''(x_0)<0$.) Then the function $f$ supposes to be strictly increasing, so $f$ is not bounded on $mathbb R$.
Then it follows the conclusion is true.
I am doubted whether my proof is correct or not, and am interested in finding a relevant example, or a counterexample for the conclusion.
derivatives proof-verification
asked yesterday
weilam06
14811
add a comment |
Ask: Does a bounded function $f$ on $mathbb R$, which is continuous and twice differentiable on $mathbb R$, has zero for its second derivative, i.e. there exists a point $x_0 in mathbb R$ such that
$$f''(x_0)=0$$
My idea is considering the contrapositive for the statement above. Let say
$$f''(x_0)>0$$
, or equivalently the function $f$ is strictly convex. (Another case: Consider the function $-f$, which is strictly concave and having $-f''(x_0)<0$.) Then the function $f$ supposes to be strictly increasing, so $f$ is not bounded on $mathbb R$.
Then it follows the conclusion is true.
I am doubted whether my proof is correct or not, and am interested in finding a relevant example, or a counterexample for the conclusion.
derivatives proof-verification
asked yesterday
weilam06
14811
add a comment |
Ask: Does a bounded function $f$ on $mathbb R$, which is continuous and twice differentiable on $mathbb R$, has zero for its second derivative, i.e. there exists a point $x_0 in mathbb R$ such that
$$f''(x_0)=0$$
My idea is considering the contrapositive for the statement above. Let say
$$f''(x_0)>0$$
, or equivalently the function $f$ is strictly convex. (Another case: Consider the function $-f$, which is strictly concave and having $-f''(x_0)<0$.) Then the function $f$ supposes to be strictly increasing, so $f$ is not bounded on $mathbb R$.
Then it follows the conclusion is true.
I am doubted whether my proof is correct or not, and am interested in finding a relevant example, or a counterexample for the conclusion.
derivatives proof-verification
asked yesterday
weilam06
14811
Ask: Does a bounded function $f$ on $mathbb R$, which is continuous and twice differentiable on $mathbb R$, has zero for its second derivative, i.e. there exists a point $x_0 in mathbb R$ such that
$$f''(x_0)=0$$
My idea is considering the contrapositive for the statement above. Let say
$$f''(x_0)>0$$
, or equivalently the function $f$ is strictly convex. (Another case: Consider the function $-f$, which is strictly concave and having $-f''(x_0)<0$.) Then the function $f$ supposes to be strictly increasing, so $f$ is not bounded on $mathbb R$.
Then it follows the conclusion is true.
I am doubted whether my proof is correct or not, and am interested in finding a relevant example, or a counterexample for the conclusion.
derivatives proof-verification
derivatives proof-verification
asked yesterday
weilam06
14811
asked yesterday
weilam06
14811
edited yesterday
asked yesterday
weilam06
14811
asked yesterday
weilam06
14811
asked yesterday
weilam06
14811
14811
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add a comment |
2 Answers
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There exists functions $f$ that are strictly increasing but bounded, take $f = arctan(x)$. So that implication in the last statement of your proof is not right.
answered yesterday
twnly
53619
add a comment |
Your idea is correct. However you need to prove why a convex map can’t be bounded.
An argument is that such a map is always above its tangent. Then consider the value of $f^prime(x)$. I can’t be always vanishing.
If $f^prime(x_0)>0$ then $limlimits_{x to infty} f(x) =infty$. And if $f^prime(x_0)<0$ then $limlimits_{x to -infty} f(x) =infty$. Due again to the fact that a convex map lies above its tangents.
answered yesterday
mathcounterexamples.net
25.2k21953
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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There exists functions $f$ that are strictly increasing but bounded, take $f = arctan(x)$. So that implication in the last statement of your proof is not right.
answered yesterday
twnly
53619
add a comment |
There exists functions $f$ that are strictly increasing but bounded, take $f = arctan(x)$. So that implication in the last statement of your proof is not right.
answered yesterday
twnly
53619
add a comment |
There exists functions $f$ that are strictly increasing but bounded, take $f = arctan(x)$. So that implication in the last statement of your proof is not right.
answered yesterday
twnly
53619
There exists functions $f$ that are strictly increasing but bounded, take $f = arctan(x)$. So that implication in the last statement of your proof is not right.
answered yesterday
twnly
53619
answered yesterday
twnly
53619
answered yesterday
twnly
53619
answered yesterday
twnly
53619
53619
add a comment |
add a comment |
Your idea is correct. However you need to prove why a convex map can’t be bounded.
An argument is that such a map is always above its tangent. Then consider the value of $f^prime(x)$. I can’t be always vanishing.
If $f^prime(x_0)>0$ then $limlimits_{x to infty} f(x) =infty$. And if $f^prime(x_0)<0$ then $limlimits_{x to -infty} f(x) =infty$. Due again to the fact that a convex map lies above its tangents.
answered yesterday
mathcounterexamples.net
25.2k21953
add a comment |
Your idea is correct. However you need to prove why a convex map can’t be bounded.
An argument is that such a map is always above its tangent. Then consider the value of $f^prime(x)$. I can’t be always vanishing.
If $f^prime(x_0)>0$ then $limlimits_{x to infty} f(x) =infty$. And if $f^prime(x_0)<0$ then $limlimits_{x to -infty} f(x) =infty$. Due again to the fact that a convex map lies above its tangents.
answered yesterday
mathcounterexamples.net
25.2k21953
add a comment |
Your idea is correct. However you need to prove why a convex map can’t be bounded.
An argument is that such a map is always above its tangent. Then consider the value of $f^prime(x)$. I can’t be always vanishing.
If $f^prime(x_0)>0$ then $limlimits_{x to infty} f(x) =infty$. And if $f^prime(x_0)<0$ then $limlimits_{x to -infty} f(x) =infty$. Due again to the fact that a convex map lies above its tangents.
answered yesterday
mathcounterexamples.net
25.2k21953
Your idea is correct. However you need to prove why a convex map can’t be bounded.
An argument is that such a map is always above its tangent. Then consider the value of $f^prime(x)$. I can’t be always vanishing.
If $f^prime(x_0)>0$ then $limlimits_{x to infty} f(x) =infty$. And if $f^prime(x_0)<0$ then $limlimits_{x to -infty} f(x) =infty$. Due again to the fact that a convex map lies above its tangents.
answered yesterday
mathcounterexamples.net
25.2k21953
answered yesterday
mathcounterexamples.net
25.2k21953
answered yesterday
mathcounterexamples.net
25.2k21953
answered yesterday
mathcounterexamples.net
25.2k21953
25.2k21953
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