Understanding lemma for proof of convexity
$begingroup$
The lemma says:
Lemma: Suppose $f$ is differentiable and $f'$ is increasing. If $a < b$ and $f(a) = f(b)$, then $f(x) < f(a) = f(b)$ for $a < x < b$.
Proof: Suppose first that $f(x) > f(a) = f(b)$ for some $x$ in $(a, b)$. Then the maximum of $f$ on $[a, b]$ occurs at some point $x_0$ in $(a, b)$ with $f(x_0) > f(a)$ and of course $f'(x_0) = 0$.
Then the proof continues:
On the other hand, applying the Mean Value Theorem to the interval $[a, x_0]$, we find that there is $x_1$ with $a < x_1 < x_0$ and
$$f'(x_1) = frac{f(x_0) - f(a)}{x_0 - a} > 0$$
Contradicting the fact that $f'$ is increasing. [...]
I understand the lemma tells us that if $f'$ is increasing, then the graph of $f$ will be below any horizontal secant line through $(a, f(a))$ and $(b, f(b))$.
I can also see the proof is by contradiction here, we will show that if $f(x) > f(a) = f(b)$ then $f'$ can't be increasing. Also clear is that $f'(x_0) = 0$ since we are looking at a local maximum for the first case of the proof.
What I'm having trouble with is what the application of the Mean Value Theorem. For the first case where we work with the local maximum, I understand we find a point $a < x_1 < x_0$ with the same slope as the line connecting the endpoints of the interval $[a, x_0]$ would have, but how does it show $f'$ can't be increasing and how do we know $f'(x_1) > 0$?
real-analysis calculus proof-explanation
asked Dec 31 '18 at 23:29
MaxMax
674722
$endgroup$
add a comment |
$begingroup$
The lemma says:
Lemma: Suppose $f$ is differentiable and $f'$ is increasing. If $a < b$ and $f(a) = f(b)$, then $f(x) < f(a) = f(b)$ for $a < x < b$.
Proof: Suppose first that $f(x) > f(a) = f(b)$ for some $x$ in $(a, b)$. Then the maximum of $f$ on $[a, b]$ occurs at some point $x_0$ in $(a, b)$ with $f(x_0) > f(a)$ and of course $f'(x_0) = 0$.
Then the proof continues:
On the other hand, applying the Mean Value Theorem to the interval $[a, x_0]$, we find that there is $x_1$ with $a < x_1 < x_0$ and
$$f'(x_1) = frac{f(x_0) - f(a)}{x_0 - a} > 0$$
Contradicting the fact that $f'$ is increasing. [...]
I understand the lemma tells us that if $f'$ is increasing, then the graph of $f$ will be below any horizontal secant line through $(a, f(a))$ and $(b, f(b))$.
I can also see the proof is by contradiction here, we will show that if $f(x) > f(a) = f(b)$ then $f'$ can't be increasing. Also clear is that $f'(x_0) = 0$ since we are looking at a local maximum for the first case of the proof.
What I'm having trouble with is what the application of the Mean Value Theorem. For the first case where we work with the local maximum, I understand we find a point $a < x_1 < x_0$ with the same slope as the line connecting the endpoints of the interval $[a, x_0]$ would have, but how does it show $f'$ can't be increasing and how do we know $f'(x_1) > 0$?
real-analysis calculus proof-explanation
asked Dec 31 '18 at 23:29
MaxMax
674722
$endgroup$
add a comment |
$begingroup$
The lemma says:
Lemma: Suppose $f$ is differentiable and $f'$ is increasing. If $a < b$ and $f(a) = f(b)$, then $f(x) < f(a) = f(b)$ for $a < x < b$.
Proof: Suppose first that $f(x) > f(a) = f(b)$ for some $x$ in $(a, b)$. Then the maximum of $f$ on $[a, b]$ occurs at some point $x_0$ in $(a, b)$ with $f(x_0) > f(a)$ and of course $f'(x_0) = 0$.
Then the proof continues:
On the other hand, applying the Mean Value Theorem to the interval $[a, x_0]$, we find that there is $x_1$ with $a < x_1 < x_0$ and
$$f'(x_1) = frac{f(x_0) - f(a)}{x_0 - a} > 0$$
Contradicting the fact that $f'$ is increasing. [...]
I understand the lemma tells us that if $f'$ is increasing, then the graph of $f$ will be below any horizontal secant line through $(a, f(a))$ and $(b, f(b))$.
I can also see the proof is by contradiction here, we will show that if $f(x) > f(a) = f(b)$ then $f'$ can't be increasing. Also clear is that $f'(x_0) = 0$ since we are looking at a local maximum for the first case of the proof.
What I'm having trouble with is what the application of the Mean Value Theorem. For the first case where we work with the local maximum, I understand we find a point $a < x_1 < x_0$ with the same slope as the line connecting the endpoints of the interval $[a, x_0]$ would have, but how does it show $f'$ can't be increasing and how do we know $f'(x_1) > 0$?
real-analysis calculus proof-explanation
asked Dec 31 '18 at 23:29
MaxMax
674722
$endgroup$
The lemma says:
Lemma: Suppose $f$ is differentiable and $f'$ is increasing. If $a < b$ and $f(a) = f(b)$, then $f(x) < f(a) = f(b)$ for $a < x < b$.
Proof: Suppose first that $f(x) > f(a) = f(b)$ for some $x$ in $(a, b)$. Then the maximum of $f$ on $[a, b]$ occurs at some point $x_0$ in $(a, b)$ with $f(x_0) > f(a)$ and of course $f'(x_0) = 0$.
Then the proof continues:
On the other hand, applying the Mean Value Theorem to the interval $[a, x_0]$, we find that there is $x_1$ with $a < x_1 < x_0$ and
$$f'(x_1) = frac{f(x_0) - f(a)}{x_0 - a} > 0$$
Contradicting the fact that $f'$ is increasing. [...]
I understand the lemma tells us that if $f'$ is increasing, then the graph of $f$ will be below any horizontal secant line through $(a, f(a))$ and $(b, f(b))$.
I can also see the proof is by contradiction here, we will show that if $f(x) > f(a) = f(b)$ then $f'$ can't be increasing. Also clear is that $f'(x_0) = 0$ since we are looking at a local maximum for the first case of the proof.
What I'm having trouble with is what the application of the Mean Value Theorem. For the first case where we work with the local maximum, I understand we find a point $a < x_1 < x_0$ with the same slope as the line connecting the endpoints of the interval $[a, x_0]$ would have, but how does it show $f'$ can't be increasing and how do we know $f'(x_1) > 0$?
real-analysis calculus proof-explanation
real-analysis calculus proof-explanation
asked Dec 31 '18 at 23:29
MaxMax
674722
asked Dec 31 '18 at 23:29
MaxMax
674722
asked Dec 31 '18 at 23:29
MaxMax
674722
asked Dec 31 '18 at 23:29
MaxMax
674722
asked Dec 31 '18 at 23:29
MaxMax
674722
674722
add a comment |
add a comment |
1 Answer
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First, it shows that $f'$ can't always be increasing because it is positive at $x_1$ but it's $0$ at $x_0 gt x_1$, so $f'$ must have decreased. As for knowing $f'left(x_1right) gt 0$, since the maximum of $f$ occurs at $x_0$, then $fleft(x_0right) gt fleft(aright)$. Also, by the condition for $x_0$, you have that $x_0 gt a$. Thus, both the numerator and denominator of $f'left(x_1right)$ are positive, making the result $gt 0$.
answered Dec 31 '18 at 23:37
John OmielanJohn Omielan
4,1962215
$endgroup$
$begingroup$
I think I get it a bit more, although I find the details still hard to grok. But generally, the proof relies on the fact that if $f'$ is increasing, then $f''$ is positive on the whole interval, is that right?
$endgroup$
– Max
Jan 1 at 2:40
$begingroup$
Because if we find a point where $f'$ is positive and then another point after that one where $f' = 0$, then $f'$ must have been decreasing between those two points and then $f''$ would have been negative.
$endgroup$
– Max
Jan 1 at 2:42
$begingroup$
@Max Even though $f'$ is increasing, there is no guarantee that $f''$ even exists, but it must be positive if it does. As for your second comment, you are correct as well re: $f''$, but the main issue is that $f'$ decreasing contradicts the original problem statement, so the original assumption must be incorrect, proving the desired result using proof by contradiction.
$endgroup$
– John Omielan
Jan 1 at 2:45
$begingroup$
Thanks again, i need to let it sink in, but I feel I can get it now. Cheers!
$endgroup$
– Max
Jan 1 at 2:58
$begingroup$
@Max You are welcome. I'm glad I was able to be of some help.
$endgroup$
– John Omielan
Jan 1 at 2:59
add a comment |
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$begingroup$
First, it shows that $f'$ can't always be increasing because it is positive at $x_1$ but it's $0$ at $x_0 gt x_1$, so $f'$ must have decreased. As for knowing $f'left(x_1right) gt 0$, since the maximum of $f$ occurs at $x_0$, then $fleft(x_0right) gt fleft(aright)$. Also, by the condition for $x_0$, you have that $x_0 gt a$. Thus, both the numerator and denominator of $f'left(x_1right)$ are positive, making the result $gt 0$.
answered Dec 31 '18 at 23:37
John OmielanJohn Omielan
4,1962215
$endgroup$
$begingroup$
I think I get it a bit more, although I find the details still hard to grok. But generally, the proof relies on the fact that if $f'$ is increasing, then $f''$ is positive on the whole interval, is that right?
$endgroup$
– Max
Jan 1 at 2:40
$begingroup$
Because if we find a point where $f'$ is positive and then another point after that one where $f' = 0$, then $f'$ must have been decreasing between those two points and then $f''$ would have been negative.
$endgroup$
– Max
Jan 1 at 2:42
$begingroup$
@Max Even though $f'$ is increasing, there is no guarantee that $f''$ even exists, but it must be positive if it does. As for your second comment, you are correct as well re: $f''$, but the main issue is that $f'$ decreasing contradicts the original problem statement, so the original assumption must be incorrect, proving the desired result using proof by contradiction.
$endgroup$
– John Omielan
Jan 1 at 2:45
$begingroup$
Thanks again, i need to let it sink in, but I feel I can get it now. Cheers!
$endgroup$
– Max
Jan 1 at 2:58
$begingroup$
@Max You are welcome. I'm glad I was able to be of some help.
$endgroup$
– John Omielan
Jan 1 at 2:59
add a comment |
$begingroup$
First, it shows that $f'$ can't always be increasing because it is positive at $x_1$ but it's $0$ at $x_0 gt x_1$, so $f'$ must have decreased. As for knowing $f'left(x_1right) gt 0$, since the maximum of $f$ occurs at $x_0$, then $fleft(x_0right) gt fleft(aright)$. Also, by the condition for $x_0$, you have that $x_0 gt a$. Thus, both the numerator and denominator of $f'left(x_1right)$ are positive, making the result $gt 0$.
answered Dec 31 '18 at 23:37
John OmielanJohn Omielan
4,1962215
$endgroup$
$begingroup$
I think I get it a bit more, although I find the details still hard to grok. But generally, the proof relies on the fact that if $f'$ is increasing, then $f''$ is positive on the whole interval, is that right?
$endgroup$
– Max
Jan 1 at 2:40
$begingroup$
Because if we find a point where $f'$ is positive and then another point after that one where $f' = 0$, then $f'$ must have been decreasing between those two points and then $f''$ would have been negative.
$endgroup$
– Max
Jan 1 at 2:42
$begingroup$
@Max Even though $f'$ is increasing, there is no guarantee that $f''$ even exists, but it must be positive if it does. As for your second comment, you are correct as well re: $f''$, but the main issue is that $f'$ decreasing contradicts the original problem statement, so the original assumption must be incorrect, proving the desired result using proof by contradiction.
$endgroup$
– John Omielan
Jan 1 at 2:45
$begingroup$
Thanks again, i need to let it sink in, but I feel I can get it now. Cheers!
$endgroup$
– Max
Jan 1 at 2:58
$begingroup$
@Max You are welcome. I'm glad I was able to be of some help.
$endgroup$
– John Omielan
Jan 1 at 2:59
add a comment |
$begingroup$
First, it shows that $f'$ can't always be increasing because it is positive at $x_1$ but it's $0$ at $x_0 gt x_1$, so $f'$ must have decreased. As for knowing $f'left(x_1right) gt 0$, since the maximum of $f$ occurs at $x_0$, then $fleft(x_0right) gt fleft(aright)$. Also, by the condition for $x_0$, you have that $x_0 gt a$. Thus, both the numerator and denominator of $f'left(x_1right)$ are positive, making the result $gt 0$.
answered Dec 31 '18 at 23:37
John OmielanJohn Omielan
4,1962215
$endgroup$
First, it shows that $f'$ can't always be increasing because it is positive at $x_1$ but it's $0$ at $x_0 gt x_1$, so $f'$ must have decreased. As for knowing $f'left(x_1right) gt 0$, since the maximum of $f$ occurs at $x_0$, then $fleft(x_0right) gt fleft(aright)$. Also, by the condition for $x_0$, you have that $x_0 gt a$. Thus, both the numerator and denominator of $f'left(x_1right)$ are positive, making the result $gt 0$.
answered Dec 31 '18 at 23:37
John OmielanJohn Omielan
4,1962215
edited Dec 31 '18 at 23:47
answered Dec 31 '18 at 23:37
John OmielanJohn Omielan
4,1962215
answered Dec 31 '18 at 23:37
John OmielanJohn Omielan
4,1962215
answered Dec 31 '18 at 23:37
John OmielanJohn Omielan
4,1962215
4,1962215
$begingroup$
I think I get it a bit more, although I find the details still hard to grok. But generally, the proof relies on the fact that if $f'$ is increasing, then $f''$ is positive on the whole interval, is that right?
$endgroup$
– Max
Jan 1 at 2:40
$begingroup$
Because if we find a point where $f'$ is positive and then another point after that one where $f' = 0$, then $f'$ must have been decreasing between those two points and then $f''$ would have been negative.
$endgroup$
– Max
Jan 1 at 2:42
$begingroup$
@Max Even though $f'$ is increasing, there is no guarantee that $f''$ even exists, but it must be positive if it does. As for your second comment, you are correct as well re: $f''$, but the main issue is that $f'$ decreasing contradicts the original problem statement, so the original assumption must be incorrect, proving the desired result using proof by contradiction.
$endgroup$
– John Omielan
Jan 1 at 2:45
$begingroup$
Thanks again, i need to let it sink in, but I feel I can get it now. Cheers!
$endgroup$
– Max
Jan 1 at 2:58
$begingroup$
@Max You are welcome. I'm glad I was able to be of some help.
$endgroup$
– John Omielan
Jan 1 at 2:59
add a comment |
$begingroup$
I think I get it a bit more, although I find the details still hard to grok. But generally, the proof relies on the fact that if $f'$ is increasing, then $f''$ is positive on the whole interval, is that right?
$endgroup$
– Max
Jan 1 at 2:40
$begingroup$
Because if we find a point where $f'$ is positive and then another point after that one where $f' = 0$, then $f'$ must have been decreasing between those two points and then $f''$ would have been negative.
$endgroup$
– Max
Jan 1 at 2:42
$begingroup$
@Max Even though $f'$ is increasing, there is no guarantee that $f''$ even exists, but it must be positive if it does. As for your second comment, you are correct as well re: $f''$, but the main issue is that $f'$ decreasing contradicts the original problem statement, so the original assumption must be incorrect, proving the desired result using proof by contradiction.
$endgroup$
– John Omielan
Jan 1 at 2:45
$begingroup$
Thanks again, i need to let it sink in, but I feel I can get it now. Cheers!
$endgroup$
– Max
Jan 1 at 2:58
$begingroup$
@Max You are welcome. I'm glad I was able to be of some help.
$endgroup$
– John Omielan
Jan 1 at 2:59
$begingroup$
I think I get it a bit more, although I find the details still hard to grok. But generally, the proof relies on the fact that if $f'$ is increasing, then $f''$ is positive on the whole interval, is that right?
$endgroup$
– Max
Jan 1 at 2:40
$begingroup$
I think I get it a bit more, although I find the details still hard to grok. But generally, the proof relies on the fact that if $f'$ is increasing, then $f''$ is positive on the whole interval, is that right?
$endgroup$
– Max
Jan 1 at 2:40
$begingroup$
Because if we find a point where $f'$ is positive and then another point after that one where $f' = 0$, then $f'$ must have been decreasing between those two points and then $f''$ would have been negative.
$endgroup$
– Max
Jan 1 at 2:42
$begingroup$
Because if we find a point where $f'$ is positive and then another point after that one where $f' = 0$, then $f'$ must have been decreasing between those two points and then $f''$ would have been negative.
$endgroup$
– Max
Jan 1 at 2:42
$begingroup$
@Max Even though $f'$ is increasing, there is no guarantee that $f''$ even exists, but it must be positive if it does. As for your second comment, you are correct as well re: $f''$, but the main issue is that $f'$ decreasing contradicts the original problem statement, so the original assumption must be incorrect, proving the desired result using proof by contradiction.
$endgroup$
– John Omielan
Jan 1 at 2:45
$begingroup$
@Max Even though $f'$ is increasing, there is no guarantee that $f''$ even exists, but it must be positive if it does. As for your second comment, you are correct as well re: $f''$, but the main issue is that $f'$ decreasing contradicts the original problem statement, so the original assumption must be incorrect, proving the desired result using proof by contradiction.
$endgroup$
– John Omielan
Jan 1 at 2:45
$begingroup$
Thanks again, i need to let it sink in, but I feel I can get it now. Cheers!
$endgroup$
– Max
Jan 1 at 2:58
$begingroup$
Thanks again, i need to let it sink in, but I feel I can get it now. Cheers!
$endgroup$
– Max
Jan 1 at 2:58
$begingroup$
@Max You are welcome. I'm glad I was able to be of some help.
$endgroup$
– John Omielan
Jan 1 at 2:59
$begingroup$
@Max You are welcome. I'm glad I was able to be of some help.
$endgroup$
– John Omielan
Jan 1 at 2:59
add a comment |
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