A finite convex function in $(a,b)$ is continuous. Moreover, $phi'$ exists except at most in a countable set and is monotone increasing
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
If $phi$ is finite and convex in $(a, b)$, then $phi$ is continuous in $(a, b)$. Moreover, $phi'$ exists except at most in a countable set and is monotone increasing.
Theorem 2.8 Every function of bounded variation has at most a countable number of discontinuities, and they are all of the first kind (jump or removable discontinuities).
- How does $(7.41)$ show in particular that $phi$ is continuous in $(a,b)$? Don't we need to show $D^-phi(x) = D^+phi(x)$ as $hto 0+$? The book shows $D^-phi(x) = D^+phi(x)$ at the end of the proof.
real-analysis analysis proof-explanation
asked Aug 10 at 22:08
user398843
399215
add a comment |Â
up vote
1
down vote
favorite
If $phi$ is finite and convex in $(a, b)$, then $phi$ is continuous in $(a, b)$. Moreover, $phi'$ exists except at most in a countable set and is monotone increasing.
Theorem 2.8 Every function of bounded variation has at most a countable number of discontinuities, and they are all of the first kind (jump or removable discontinuities).
- How does $(7.41)$ show in particular that $phi$ is continuous in $(a,b)$? Don't we need to show $D^-phi(x) = D^+phi(x)$ as $hto 0+$? The book shows $D^-phi(x) = D^+phi(x)$ at the end of the proof.
real-analysis analysis proof-explanation
asked Aug 10 at 22:08
user398843
399215
@mfl This is Theorem 7.40 on page 157 of Measure and integral by Wheeden and Zygmund. The theorem on page 157 doesn't state that $phi$ is finite, but since they say $D^+phi$ is of bounded variation I looked up the beginning of this section on page 154, and they suppose $phi$ is finite on an open interval $(a,b)$. Also see the comment by Michael Grant on this post math.stackexchange.com/q/258511/426645.
– user398843
Aug 10 at 22:32
@mfl The interval here is open, the one in the counterexample is compact, and the compactness plays a role there.
– amsmath
Aug 10 at 22:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $phi$ is finite and convex in $(a, b)$, then $phi$ is continuous in $(a, b)$. Moreover, $phi'$ exists except at most in a countable set and is monotone increasing.
Theorem 2.8 Every function of bounded variation has at most a countable number of discontinuities, and they are all of the first kind (jump or removable discontinuities).
- How does $(7.41)$ show in particular that $phi$ is continuous in $(a,b)$? Don't we need to show $D^-phi(x) = D^+phi(x)$ as $hto 0+$? The book shows $D^-phi(x) = D^+phi(x)$ at the end of the proof.
real-analysis analysis proof-explanation
asked Aug 10 at 22:08
user398843
399215
If $phi$ is finite and convex in $(a, b)$, then $phi$ is continuous in $(a, b)$. Moreover, $phi'$ exists except at most in a countable set and is monotone increasing.
Theorem 2.8 Every function of bounded variation has at most a countable number of discontinuities, and they are all of the first kind (jump or removable discontinuities).
- How does $(7.41)$ show in particular that $phi$ is continuous in $(a,b)$? Don't we need to show $D^-phi(x) = D^+phi(x)$ as $hto 0+$? The book shows $D^-phi(x) = D^+phi(x)$ at the end of the proof.
real-analysis analysis proof-explanation
asked Aug 10 at 22:08
user398843
399215
edited Aug 10 at 22:21
asked Aug 10 at 22:08
user398843
399215
asked Aug 10 at 22:08
user398843
399215
asked Aug 10 at 22:08
user398843
399215
399215
@mfl This is Theorem 7.40 on page 157 of Measure and integral by Wheeden and Zygmund. The theorem on page 157 doesn't state that $phi$ is finite, but since they say $D^+phi$ is of bounded variation I looked up the beginning of this section on page 154, and they suppose $phi$ is finite on an open interval $(a,b)$. Also see the comment by Michael Grant on this post math.stackexchange.com/q/258511/426645.
– user398843
Aug 10 at 22:32
@mfl The interval here is open, the one in the counterexample is compact, and the compactness plays a role there.
– amsmath
Aug 10 at 22:37
add a comment |Â
@mfl This is Theorem 7.40 on page 157 of Measure and integral by Wheeden and Zygmund. The theorem on page 157 doesn't state that $phi$ is finite, but since they say $D^+phi$ is of bounded variation I looked up the beginning of this section on page 154, and they suppose $phi$ is finite on an open interval $(a,b)$. Also see the comment by Michael Grant on this post math.stackexchange.com/q/258511/426645.
– user398843
Aug 10 at 22:32
@mfl The interval here is open, the one in the counterexample is compact, and the compactness plays a role there.
– amsmath
Aug 10 at 22:37
@mfl This is Theorem 7.40 on page 157 of Measure and integral by Wheeden and Zygmund. The theorem on page 157 doesn't state that $phi$ is finite, but since they say $D^+phi$ is of bounded variation I looked up the beginning of this section on page 154, and they suppose $phi$ is finite on an open interval $(a,b)$. Also see the comment by Michael Grant on this post math.stackexchange.com/q/258511/426645.
– user398843
Aug 10 at 22:32
@mfl This is Theorem 7.40 on page 157 of Measure and integral by Wheeden and Zygmund. The theorem on page 157 doesn't state that $phi$ is finite, but since they say $D^+phi$ is of bounded variation I looked up the beginning of this section on page 154, and they suppose $phi$ is finite on an open interval $(a,b)$. Also see the comment by Michael Grant on this post math.stackexchange.com/q/258511/426645.
– user398843
Aug 10 at 22:32
@mfl The interval here is open, the one in the counterexample is compact, and the compactness plays a role there.
– amsmath
Aug 10 at 22:37
@mfl The interval here is open, the one in the counterexample is compact, and the compactness plays a role there.
– amsmath
Aug 10 at 22:37
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
The existence of the limit $D^+phi(x)$ tells us in particular that $phi(x+h)tophi(x)$ as $hto 0+$. Similarly, the existence of $D^-phi(x)$ implies $phi(x-h)tophi(x)$ as $hto 0+$. These together yield that $phi$ is continuous at $x$.
answered Aug 10 at 22:43
amsmath
1,622114
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The existence of the limit $D^+phi(x)$ tells us in particular that $phi(x+h)tophi(x)$ as $hto 0+$. Similarly, the existence of $D^-phi(x)$ implies $phi(x-h)tophi(x)$ as $hto 0+$. These together yield that $phi$ is continuous at $x$.
answered Aug 10 at 22:43
amsmath
1,622114
add a comment |Â
up vote
0
down vote
accepted
The existence of the limit $D^+phi(x)$ tells us in particular that $phi(x+h)tophi(x)$ as $hto 0+$. Similarly, the existence of $D^-phi(x)$ implies $phi(x-h)tophi(x)$ as $hto 0+$. These together yield that $phi$ is continuous at $x$.
answered Aug 10 at 22:43
amsmath
1,622114
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The existence of the limit $D^+phi(x)$ tells us in particular that $phi(x+h)tophi(x)$ as $hto 0+$. Similarly, the existence of $D^-phi(x)$ implies $phi(x-h)tophi(x)$ as $hto 0+$. These together yield that $phi$ is continuous at $x$.
answered Aug 10 at 22:43
amsmath
1,622114
The existence of the limit $D^+phi(x)$ tells us in particular that $phi(x+h)tophi(x)$ as $hto 0+$. Similarly, the existence of $D^-phi(x)$ implies $phi(x-h)tophi(x)$ as $hto 0+$. These together yield that $phi$ is continuous at $x$.
answered Aug 10 at 22:43
amsmath
1,622114
answered Aug 10 at 22:43
amsmath
1,622114
answered Aug 10 at 22:43
amsmath
1,622114
answered Aug 10 at 22:43
amsmath
1,622114
1,622114
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2878859%2fa-finite-convex-function-in-a-b-is-continuous-moreover-phi-exists-exce%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
@mfl This is Theorem 7.40 on page 157 of Measure and integral by Wheeden and Zygmund. The theorem on page 157 doesn't state that $phi$ is finite, but since they say $D^+phi$ is of bounded variation I looked up the beginning of this section on page 154, and they suppose $phi$ is finite on an open interval $(a,b)$. Also see the comment by Michael Grant on this post math.stackexchange.com/q/258511/426645.
– user398843
Aug 10 at 22:32
@mfl The interval here is open, the one in the counterexample is compact, and the compactness plays a role there.
– amsmath
Aug 10 at 22:37