Prove $f$ is riemann integrable if $f$ is continuous on closed interval $[a,b]$. (Explanation)
$begingroup$
I was given the following proof of a theorem, quick note about notation, the fact that $f$ is riemann integrable on $[a,b]$ we denote by $finmathcal{R}([a,b])$
Theorem: Let $a,binmathbb{R},a<b$, let $f$ be continuous on closed interval $[a,b]$, then $finmathcal{R}([a,b])$
We use the following lemma:
Lemma: Let $a,binmathbb{R},a<b$ and $f$ bounded on $[a,b]$. Then the following is equivallent: $$finmathcal{R}([a,b])tag{1}$$ $$forall epsilon >0 exists sigma_{[a,b]}:overline{S}(f,sigma)-underline{S}(f,sigma)<epsilon$$
where $sigma$ is a partition of $[a,b]$. Now, i understand the proof of the lemma and I'm not giving it here.
Proof. (of the theorem)
$f$ is continuous on $[a,b]$, so $f$ is bounded on $[a,b]$, also because $f$ is bounded on bounded interval, f is uniformly continuous. Let $epsilon > 0$ be given, by assumption, we find $delta>0$ s.t. for any pair $x,yin[a,b]:|x-y|<delta Rightarrow |f(x)-f(y)|<epsilon$. Choose partitioning $sigma_{[a,b]}$ s.t. $nu(sigma)<delta$. Then by uniform continuity for each $iin{1ldots,n}$ we have that $$sup_{[x_{i-1},x_i]}fleq inf_{[x_{i-1},x_i]}f+epsilon$$
thus $$overline{S}(f,sigma)-underline{S}(f,sigma)=sum_{i=1}^n bigg(sup_{[x_{i-1},x_i]}f-inf_{[x_{i-1},x_i]}fbigg)(x_i-x_{i-1})leq sum_{i=1}^n epsilon(x_i-x_{i-1})=epsilon(b-a)$$
which by lemma shows that $finmathcal{R}([a,b])$.
where $nu(sigma)$ is the norm of the partition, that is $nu(sigma)=max{x_{i}-x_{i-1}mid iin{1,ldots,n}}$
Now, the part I don't understand is why uniform continuity implies something like $sup_{[x_{i-1},x_i]}fleq inf_{[x_{i-1},x_i]}f+epsilon$ (bolded part). Would someone explain more thoroughly, what is going on? Do they mean that by continuity $f$ attains extreme values (suprema, and infima) on each of $mathcal{I}:=[x_{i-1},x_i]$ and by $|f(x)-f(y)|<epsilon$, this results in $sup_mathcal{I} f - inf_mathcal{I} f leq epsilon$ (now, why is there a $leq$ instead of $<$ ?).
real-analysis integration proof-explanation riemann-integration
asked Dec 24 '18 at 1:30
Michal DvořákMichal Dvořák
966416
$endgroup$
add a comment |
$begingroup$
I was given the following proof of a theorem, quick note about notation, the fact that $f$ is riemann integrable on $[a,b]$ we denote by $finmathcal{R}([a,b])$
Theorem: Let $a,binmathbb{R},a<b$, let $f$ be continuous on closed interval $[a,b]$, then $finmathcal{R}([a,b])$
We use the following lemma:
Lemma: Let $a,binmathbb{R},a<b$ and $f$ bounded on $[a,b]$. Then the following is equivallent: $$finmathcal{R}([a,b])tag{1}$$ $$forall epsilon >0 exists sigma_{[a,b]}:overline{S}(f,sigma)-underline{S}(f,sigma)<epsilon$$
where $sigma$ is a partition of $[a,b]$. Now, i understand the proof of the lemma and I'm not giving it here.
Proof. (of the theorem)
$f$ is continuous on $[a,b]$, so $f$ is bounded on $[a,b]$, also because $f$ is bounded on bounded interval, f is uniformly continuous. Let $epsilon > 0$ be given, by assumption, we find $delta>0$ s.t. for any pair $x,yin[a,b]:|x-y|<delta Rightarrow |f(x)-f(y)|<epsilon$. Choose partitioning $sigma_{[a,b]}$ s.t. $nu(sigma)<delta$. Then by uniform continuity for each $iin{1ldots,n}$ we have that $$sup_{[x_{i-1},x_i]}fleq inf_{[x_{i-1},x_i]}f+epsilon$$
thus $$overline{S}(f,sigma)-underline{S}(f,sigma)=sum_{i=1}^n bigg(sup_{[x_{i-1},x_i]}f-inf_{[x_{i-1},x_i]}fbigg)(x_i-x_{i-1})leq sum_{i=1}^n epsilon(x_i-x_{i-1})=epsilon(b-a)$$
which by lemma shows that $finmathcal{R}([a,b])$.
where $nu(sigma)$ is the norm of the partition, that is $nu(sigma)=max{x_{i}-x_{i-1}mid iin{1,ldots,n}}$
Now, the part I don't understand is why uniform continuity implies something like $sup_{[x_{i-1},x_i]}fleq inf_{[x_{i-1},x_i]}f+epsilon$ (bolded part). Would someone explain more thoroughly, what is going on? Do they mean that by continuity $f$ attains extreme values (suprema, and infima) on each of $mathcal{I}:=[x_{i-1},x_i]$ and by $|f(x)-f(y)|<epsilon$, this results in $sup_mathcal{I} f - inf_mathcal{I} f leq epsilon$ (now, why is there a $leq$ instead of $<$ ?).
real-analysis integration proof-explanation riemann-integration
asked Dec 24 '18 at 1:30
Michal DvořákMichal Dvořák
966416
$endgroup$
add a comment |
$begingroup$
I was given the following proof of a theorem, quick note about notation, the fact that $f$ is riemann integrable on $[a,b]$ we denote by $finmathcal{R}([a,b])$
Theorem: Let $a,binmathbb{R},a<b$, let $f$ be continuous on closed interval $[a,b]$, then $finmathcal{R}([a,b])$
We use the following lemma:
Lemma: Let $a,binmathbb{R},a<b$ and $f$ bounded on $[a,b]$. Then the following is equivallent: $$finmathcal{R}([a,b])tag{1}$$ $$forall epsilon >0 exists sigma_{[a,b]}:overline{S}(f,sigma)-underline{S}(f,sigma)<epsilon$$
where $sigma$ is a partition of $[a,b]$. Now, i understand the proof of the lemma and I'm not giving it here.
Proof. (of the theorem)
$f$ is continuous on $[a,b]$, so $f$ is bounded on $[a,b]$, also because $f$ is bounded on bounded interval, f is uniformly continuous. Let $epsilon > 0$ be given, by assumption, we find $delta>0$ s.t. for any pair $x,yin[a,b]:|x-y|<delta Rightarrow |f(x)-f(y)|<epsilon$. Choose partitioning $sigma_{[a,b]}$ s.t. $nu(sigma)<delta$. Then by uniform continuity for each $iin{1ldots,n}$ we have that $$sup_{[x_{i-1},x_i]}fleq inf_{[x_{i-1},x_i]}f+epsilon$$
thus $$overline{S}(f,sigma)-underline{S}(f,sigma)=sum_{i=1}^n bigg(sup_{[x_{i-1},x_i]}f-inf_{[x_{i-1},x_i]}fbigg)(x_i-x_{i-1})leq sum_{i=1}^n epsilon(x_i-x_{i-1})=epsilon(b-a)$$
which by lemma shows that $finmathcal{R}([a,b])$.
where $nu(sigma)$ is the norm of the partition, that is $nu(sigma)=max{x_{i}-x_{i-1}mid iin{1,ldots,n}}$
Now, the part I don't understand is why uniform continuity implies something like $sup_{[x_{i-1},x_i]}fleq inf_{[x_{i-1},x_i]}f+epsilon$ (bolded part). Would someone explain more thoroughly, what is going on? Do they mean that by continuity $f$ attains extreme values (suprema, and infima) on each of $mathcal{I}:=[x_{i-1},x_i]$ and by $|f(x)-f(y)|<epsilon$, this results in $sup_mathcal{I} f - inf_mathcal{I} f leq epsilon$ (now, why is there a $leq$ instead of $<$ ?).
real-analysis integration proof-explanation riemann-integration
asked Dec 24 '18 at 1:30
Michal DvořákMichal Dvořák
966416
$endgroup$
I was given the following proof of a theorem, quick note about notation, the fact that $f$ is riemann integrable on $[a,b]$ we denote by $finmathcal{R}([a,b])$
Theorem: Let $a,binmathbb{R},a<b$, let $f$ be continuous on closed interval $[a,b]$, then $finmathcal{R}([a,b])$
We use the following lemma:
Lemma: Let $a,binmathbb{R},a<b$ and $f$ bounded on $[a,b]$. Then the following is equivallent: $$finmathcal{R}([a,b])tag{1}$$ $$forall epsilon >0 exists sigma_{[a,b]}:overline{S}(f,sigma)-underline{S}(f,sigma)<epsilon$$
where $sigma$ is a partition of $[a,b]$. Now, i understand the proof of the lemma and I'm not giving it here.
Proof. (of the theorem)
$f$ is continuous on $[a,b]$, so $f$ is bounded on $[a,b]$, also because $f$ is bounded on bounded interval, f is uniformly continuous. Let $epsilon > 0$ be given, by assumption, we find $delta>0$ s.t. for any pair $x,yin[a,b]:|x-y|<delta Rightarrow |f(x)-f(y)|<epsilon$. Choose partitioning $sigma_{[a,b]}$ s.t. $nu(sigma)<delta$. Then by uniform continuity for each $iin{1ldots,n}$ we have that $$sup_{[x_{i-1},x_i]}fleq inf_{[x_{i-1},x_i]}f+epsilon$$
thus $$overline{S}(f,sigma)-underline{S}(f,sigma)=sum_{i=1}^n bigg(sup_{[x_{i-1},x_i]}f-inf_{[x_{i-1},x_i]}fbigg)(x_i-x_{i-1})leq sum_{i=1}^n epsilon(x_i-x_{i-1})=epsilon(b-a)$$
which by lemma shows that $finmathcal{R}([a,b])$.
where $nu(sigma)$ is the norm of the partition, that is $nu(sigma)=max{x_{i}-x_{i-1}mid iin{1,ldots,n}}$
Now, the part I don't understand is why uniform continuity implies something like $sup_{[x_{i-1},x_i]}fleq inf_{[x_{i-1},x_i]}f+epsilon$ (bolded part). Would someone explain more thoroughly, what is going on? Do they mean that by continuity $f$ attains extreme values (suprema, and infima) on each of $mathcal{I}:=[x_{i-1},x_i]$ and by $|f(x)-f(y)|<epsilon$, this results in $sup_mathcal{I} f - inf_mathcal{I} f leq epsilon$ (now, why is there a $leq$ instead of $<$ ?).
real-analysis integration proof-explanation riemann-integration
real-analysis integration proof-explanation riemann-integration
asked Dec 24 '18 at 1:30
Michal DvořákMichal Dvořák
966416
asked Dec 24 '18 at 1:30
Michal DvořákMichal Dvořák
966416
edited Dec 24 '18 at 1:39
Michal Dvořák
asked Dec 24 '18 at 1:30
Michal DvořákMichal Dvořák
966416
asked Dec 24 '18 at 1:30
Michal DvořákMichal Dvořák
966416
asked Dec 24 '18 at 1:30
Michal DvořákMichal Dvořák
966416
966416
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $|s-r| <delta$, and $|f(x)-f(y)| < epsilon$ whenever $|x-y| <delta$. Then, on the interval $[r,s]$ (in the domain of $f$),
$$epsilon ge sup_{x,yin [r,s]} (f(x)-f(y)) =sup_{xin [r,s]}f(x) - inf_{xin [r,s]}f(y)$$
That equality is always true, even without continuity; the supremum minus the infimum is an upper bound for the variation, and we can approach arbitrarily close to each of them. What (uniform) continuity gets us is the inequality out front relating to $epsilon$, as we can choose the mesh fine enough that the variation in any one subinterval is small.
answered Dec 24 '18 at 1:51
jmerryjmerry
9,7681225
$endgroup$
$begingroup$
Oh, I see now, also, we can interchange, by symmetry of $x,y$ on $[a,b]$ right? So maybe this could also be $sup_{x,yin I}(f(x)-f(y))=sup_{x,yin I}(f(y)-f(x))=sup_{xin I} f(x) - inf_{yin I}f(y)=$... my argument would go as follows: by boundeness suprema $sup f =: f(M_i)$ and infima $inf f =: f(m_i)$exist on each of $[x_{i-1},x_i]$ and by the continuity $|f(M_i)-f(m_i)| = f(M_i)-f(m_i) < epsilon$, would this also be a good reasoning?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:02
$begingroup$
I just changed the variable names - my interval is not the same as your $[a,b]$, and now it uses different letters. This argument is all running on one of the small intervals belonging to the partition. Now, your argument, calling the supremum $f(M_i)$? That invokes continuity; you're saying that the function has a maximum, not just a supremum. Be careful with that sort of thing.
$endgroup$
– jmerry
Dec 24 '18 at 2:11
$begingroup$
Well, then we use the continuity again to show that infima are in fact minima and suprema are maxima on the interval, or?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:15
$begingroup$
yes, you can, but its not really relevant to the argument. Saying $sup f = f(M_i)$ is technically correct by continuity on the closed interval, but you may not always have $M_i$ which allows for $f(M_i) = sup_{xin I}f(x)$
$endgroup$
– rubikscube09
Dec 24 '18 at 3:20
add a comment |
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$begingroup$
Suppose that $|s-r| <delta$, and $|f(x)-f(y)| < epsilon$ whenever $|x-y| <delta$. Then, on the interval $[r,s]$ (in the domain of $f$),
$$epsilon ge sup_{x,yin [r,s]} (f(x)-f(y)) =sup_{xin [r,s]}f(x) - inf_{xin [r,s]}f(y)$$
That equality is always true, even without continuity; the supremum minus the infimum is an upper bound for the variation, and we can approach arbitrarily close to each of them. What (uniform) continuity gets us is the inequality out front relating to $epsilon$, as we can choose the mesh fine enough that the variation in any one subinterval is small.
answered Dec 24 '18 at 1:51
jmerryjmerry
9,7681225
$endgroup$
$begingroup$
Oh, I see now, also, we can interchange, by symmetry of $x,y$ on $[a,b]$ right? So maybe this could also be $sup_{x,yin I}(f(x)-f(y))=sup_{x,yin I}(f(y)-f(x))=sup_{xin I} f(x) - inf_{yin I}f(y)=$... my argument would go as follows: by boundeness suprema $sup f =: f(M_i)$ and infima $inf f =: f(m_i)$exist on each of $[x_{i-1},x_i]$ and by the continuity $|f(M_i)-f(m_i)| = f(M_i)-f(m_i) < epsilon$, would this also be a good reasoning?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:02
$begingroup$
I just changed the variable names - my interval is not the same as your $[a,b]$, and now it uses different letters. This argument is all running on one of the small intervals belonging to the partition. Now, your argument, calling the supremum $f(M_i)$? That invokes continuity; you're saying that the function has a maximum, not just a supremum. Be careful with that sort of thing.
$endgroup$
– jmerry
Dec 24 '18 at 2:11
$begingroup$
Well, then we use the continuity again to show that infima are in fact minima and suprema are maxima on the interval, or?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:15
$begingroup$
yes, you can, but its not really relevant to the argument. Saying $sup f = f(M_i)$ is technically correct by continuity on the closed interval, but you may not always have $M_i$ which allows for $f(M_i) = sup_{xin I}f(x)$
$endgroup$
– rubikscube09
Dec 24 '18 at 3:20
add a comment |
$begingroup$
Suppose that $|s-r| <delta$, and $|f(x)-f(y)| < epsilon$ whenever $|x-y| <delta$. Then, on the interval $[r,s]$ (in the domain of $f$),
$$epsilon ge sup_{x,yin [r,s]} (f(x)-f(y)) =sup_{xin [r,s]}f(x) - inf_{xin [r,s]}f(y)$$
That equality is always true, even without continuity; the supremum minus the infimum is an upper bound for the variation, and we can approach arbitrarily close to each of them. What (uniform) continuity gets us is the inequality out front relating to $epsilon$, as we can choose the mesh fine enough that the variation in any one subinterval is small.
answered Dec 24 '18 at 1:51
jmerryjmerry
9,7681225
$endgroup$
$begingroup$
Oh, I see now, also, we can interchange, by symmetry of $x,y$ on $[a,b]$ right? So maybe this could also be $sup_{x,yin I}(f(x)-f(y))=sup_{x,yin I}(f(y)-f(x))=sup_{xin I} f(x) - inf_{yin I}f(y)=$... my argument would go as follows: by boundeness suprema $sup f =: f(M_i)$ and infima $inf f =: f(m_i)$exist on each of $[x_{i-1},x_i]$ and by the continuity $|f(M_i)-f(m_i)| = f(M_i)-f(m_i) < epsilon$, would this also be a good reasoning?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:02
$begingroup$
I just changed the variable names - my interval is not the same as your $[a,b]$, and now it uses different letters. This argument is all running on one of the small intervals belonging to the partition. Now, your argument, calling the supremum $f(M_i)$? That invokes continuity; you're saying that the function has a maximum, not just a supremum. Be careful with that sort of thing.
$endgroup$
– jmerry
Dec 24 '18 at 2:11
$begingroup$
Well, then we use the continuity again to show that infima are in fact minima and suprema are maxima on the interval, or?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:15
$begingroup$
yes, you can, but its not really relevant to the argument. Saying $sup f = f(M_i)$ is technically correct by continuity on the closed interval, but you may not always have $M_i$ which allows for $f(M_i) = sup_{xin I}f(x)$
$endgroup$
– rubikscube09
Dec 24 '18 at 3:20
add a comment |
$begingroup$
Suppose that $|s-r| <delta$, and $|f(x)-f(y)| < epsilon$ whenever $|x-y| <delta$. Then, on the interval $[r,s]$ (in the domain of $f$),
$$epsilon ge sup_{x,yin [r,s]} (f(x)-f(y)) =sup_{xin [r,s]}f(x) - inf_{xin [r,s]}f(y)$$
That equality is always true, even without continuity; the supremum minus the infimum is an upper bound for the variation, and we can approach arbitrarily close to each of them. What (uniform) continuity gets us is the inequality out front relating to $epsilon$, as we can choose the mesh fine enough that the variation in any one subinterval is small.
answered Dec 24 '18 at 1:51
jmerryjmerry
9,7681225
$endgroup$
Suppose that $|s-r| <delta$, and $|f(x)-f(y)| < epsilon$ whenever $|x-y| <delta$. Then, on the interval $[r,s]$ (in the domain of $f$),
$$epsilon ge sup_{x,yin [r,s]} (f(x)-f(y)) =sup_{xin [r,s]}f(x) - inf_{xin [r,s]}f(y)$$
That equality is always true, even without continuity; the supremum minus the infimum is an upper bound for the variation, and we can approach arbitrarily close to each of them. What (uniform) continuity gets us is the inequality out front relating to $epsilon$, as we can choose the mesh fine enough that the variation in any one subinterval is small.
answered Dec 24 '18 at 1:51
jmerryjmerry
9,7681225
edited Dec 24 '18 at 2:05
answered Dec 24 '18 at 1:51
jmerryjmerry
9,7681225
answered Dec 24 '18 at 1:51
jmerryjmerry
9,7681225
answered Dec 24 '18 at 1:51
jmerryjmerry
9,7681225
9,7681225
$begingroup$
Oh, I see now, also, we can interchange, by symmetry of $x,y$ on $[a,b]$ right? So maybe this could also be $sup_{x,yin I}(f(x)-f(y))=sup_{x,yin I}(f(y)-f(x))=sup_{xin I} f(x) - inf_{yin I}f(y)=$... my argument would go as follows: by boundeness suprema $sup f =: f(M_i)$ and infima $inf f =: f(m_i)$exist on each of $[x_{i-1},x_i]$ and by the continuity $|f(M_i)-f(m_i)| = f(M_i)-f(m_i) < epsilon$, would this also be a good reasoning?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:02
$begingroup$
I just changed the variable names - my interval is not the same as your $[a,b]$, and now it uses different letters. This argument is all running on one of the small intervals belonging to the partition. Now, your argument, calling the supremum $f(M_i)$? That invokes continuity; you're saying that the function has a maximum, not just a supremum. Be careful with that sort of thing.
$endgroup$
– jmerry
Dec 24 '18 at 2:11
$begingroup$
Well, then we use the continuity again to show that infima are in fact minima and suprema are maxima on the interval, or?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:15
$begingroup$
yes, you can, but its not really relevant to the argument. Saying $sup f = f(M_i)$ is technically correct by continuity on the closed interval, but you may not always have $M_i$ which allows for $f(M_i) = sup_{xin I}f(x)$
$endgroup$
– rubikscube09
Dec 24 '18 at 3:20
add a comment |
$begingroup$
Oh, I see now, also, we can interchange, by symmetry of $x,y$ on $[a,b]$ right? So maybe this could also be $sup_{x,yin I}(f(x)-f(y))=sup_{x,yin I}(f(y)-f(x))=sup_{xin I} f(x) - inf_{yin I}f(y)=$... my argument would go as follows: by boundeness suprema $sup f =: f(M_i)$ and infima $inf f =: f(m_i)$exist on each of $[x_{i-1},x_i]$ and by the continuity $|f(M_i)-f(m_i)| = f(M_i)-f(m_i) < epsilon$, would this also be a good reasoning?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:02
$begingroup$
I just changed the variable names - my interval is not the same as your $[a,b]$, and now it uses different letters. This argument is all running on one of the small intervals belonging to the partition. Now, your argument, calling the supremum $f(M_i)$? That invokes continuity; you're saying that the function has a maximum, not just a supremum. Be careful with that sort of thing.
$endgroup$
– jmerry
Dec 24 '18 at 2:11
$begingroup$
Well, then we use the continuity again to show that infima are in fact minima and suprema are maxima on the interval, or?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:15
$begingroup$
yes, you can, but its not really relevant to the argument. Saying $sup f = f(M_i)$ is technically correct by continuity on the closed interval, but you may not always have $M_i$ which allows for $f(M_i) = sup_{xin I}f(x)$
$endgroup$
– rubikscube09
Dec 24 '18 at 3:20
$begingroup$
Oh, I see now, also, we can interchange, by symmetry of $x,y$ on $[a,b]$ right? So maybe this could also be $sup_{x,yin I}(f(x)-f(y))=sup_{x,yin I}(f(y)-f(x))=sup_{xin I} f(x) - inf_{yin I}f(y)=$... my argument would go as follows: by boundeness suprema $sup f =: f(M_i)$ and infima $inf f =: f(m_i)$exist on each of $[x_{i-1},x_i]$ and by the continuity $|f(M_i)-f(m_i)| = f(M_i)-f(m_i) < epsilon$, would this also be a good reasoning?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:02
$begingroup$
Oh, I see now, also, we can interchange, by symmetry of $x,y$ on $[a,b]$ right? So maybe this could also be $sup_{x,yin I}(f(x)-f(y))=sup_{x,yin I}(f(y)-f(x))=sup_{xin I} f(x) - inf_{yin I}f(y)=$... my argument would go as follows: by boundeness suprema $sup f =: f(M_i)$ and infima $inf f =: f(m_i)$exist on each of $[x_{i-1},x_i]$ and by the continuity $|f(M_i)-f(m_i)| = f(M_i)-f(m_i) < epsilon$, would this also be a good reasoning?
$endgroup$
– Michal Dvořák
Dec 24 '18 at 2:02
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I just changed the variable names - my interval is not the same as your $[a,b]$, and now it uses different letters. This argument is all running on one of the small intervals belonging to the partition. Now, your argument, calling the supremum $f(M_i)$? That invokes continuity; you're saying that the function has a maximum, not just a supremum. Be careful with that sort of thing.
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– jmerry
Dec 24 '18 at 2:11
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I just changed the variable names - my interval is not the same as your $[a,b]$, and now it uses different letters. This argument is all running on one of the small intervals belonging to the partition. Now, your argument, calling the supremum $f(M_i)$? That invokes continuity; you're saying that the function has a maximum, not just a supremum. Be careful with that sort of thing.
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– jmerry
Dec 24 '18 at 2:11
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Well, then we use the continuity again to show that infima are in fact minima and suprema are maxima on the interval, or?
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– Michal Dvořák
Dec 24 '18 at 2:15
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Well, then we use the continuity again to show that infima are in fact minima and suprema are maxima on the interval, or?
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– Michal Dvořák
Dec 24 '18 at 2:15
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yes, you can, but its not really relevant to the argument. Saying $sup f = f(M_i)$ is technically correct by continuity on the closed interval, but you may not always have $M_i$ which allows for $f(M_i) = sup_{xin I}f(x)$
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– rubikscube09
Dec 24 '18 at 3:20
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yes, you can, but its not really relevant to the argument. Saying $sup f = f(M_i)$ is technically correct by continuity on the closed interval, but you may not always have $M_i$ which allows for $f(M_i) = sup_{xin I}f(x)$
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– rubikscube09
Dec 24 '18 at 3:20
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