If the integral is positive then show that there is a subinterval and $m>0$ such that $f(x)ge m$ on the...
$begingroup$
Suppose $int_{a}^{b} f$ exists and is positive. Prove that there
exists a subinterval $Jsubset [a,b]$ and a constant $m>0$ such that
$f(x)ge m$ for $xin J$.
The above question is from my analysis textbook and here's my attempt at proving it.
If $int_{a}^{b} f$ exists then $int_{a}^{b} f = underline{int_{a}^{b} f}$ by definition. By definition
$$underline{int_{a}^{b} f} = sup { L(P,f) mid text{P is a partition of [a,b]}}$$
Since $0$ is not an upperbound for the above set (inside the $sup$), There must be a partition $P$ such that $L(P,f) >0$. By definition,
$$L(P,f)= sum_{k=1}^{n} m_{k} (f) Delta x_k$$
If $m_k (f)$ were less than or equal to $0$ for each $k$ then $L(P,f) le 0$, thus, it must be that $m_k (f) > 0$ for some $k$. We know that $m_{k} (f) = inf { f(x) mid xin [x_{k-1} , x_{k} ] }$. Thus, we let $m=m_{k} (f)$ and it follows that $f(x)ge m$ for $xin [x_{k-1} , x_{k}]$.
Is my proof okay? Is there any easier way through this?
real-analysis proof-verification proof-writing alternative-proof riemann-integration
asked Jan 4 at 10:24
Ashish KAshish K
910613
$endgroup$
add a comment |
$begingroup$
Suppose $int_{a}^{b} f$ exists and is positive. Prove that there
exists a subinterval $Jsubset [a,b]$ and a constant $m>0$ such that
$f(x)ge m$ for $xin J$.
The above question is from my analysis textbook and here's my attempt at proving it.
If $int_{a}^{b} f$ exists then $int_{a}^{b} f = underline{int_{a}^{b} f}$ by definition. By definition
$$underline{int_{a}^{b} f} = sup { L(P,f) mid text{P is a partition of [a,b]}}$$
Since $0$ is not an upperbound for the above set (inside the $sup$), There must be a partition $P$ such that $L(P,f) >0$. By definition,
$$L(P,f)= sum_{k=1}^{n} m_{k} (f) Delta x_k$$
If $m_k (f)$ were less than or equal to $0$ for each $k$ then $L(P,f) le 0$, thus, it must be that $m_k (f) > 0$ for some $k$. We know that $m_{k} (f) = inf { f(x) mid xin [x_{k-1} , x_{k} ] }$. Thus, we let $m=m_{k} (f)$ and it follows that $f(x)ge m$ for $xin [x_{k-1} , x_{k}]$.
Is my proof okay? Is there any easier way through this?
real-analysis proof-verification proof-writing alternative-proof riemann-integration
asked Jan 4 at 10:24
Ashish KAshish K
910613
$endgroup$
2
$begingroup$
Your proof is fine and there is no better way.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 10:29
add a comment |
$begingroup$
Suppose $int_{a}^{b} f$ exists and is positive. Prove that there
exists a subinterval $Jsubset [a,b]$ and a constant $m>0$ such that
$f(x)ge m$ for $xin J$.
The above question is from my analysis textbook and here's my attempt at proving it.
If $int_{a}^{b} f$ exists then $int_{a}^{b} f = underline{int_{a}^{b} f}$ by definition. By definition
$$underline{int_{a}^{b} f} = sup { L(P,f) mid text{P is a partition of [a,b]}}$$
Since $0$ is not an upperbound for the above set (inside the $sup$), There must be a partition $P$ such that $L(P,f) >0$. By definition,
$$L(P,f)= sum_{k=1}^{n} m_{k} (f) Delta x_k$$
If $m_k (f)$ were less than or equal to $0$ for each $k$ then $L(P,f) le 0$, thus, it must be that $m_k (f) > 0$ for some $k$. We know that $m_{k} (f) = inf { f(x) mid xin [x_{k-1} , x_{k} ] }$. Thus, we let $m=m_{k} (f)$ and it follows that $f(x)ge m$ for $xin [x_{k-1} , x_{k}]$.
Is my proof okay? Is there any easier way through this?
real-analysis proof-verification proof-writing alternative-proof riemann-integration
asked Jan 4 at 10:24
Ashish KAshish K
910613
$endgroup$
Suppose $int_{a}^{b} f$ exists and is positive. Prove that there
exists a subinterval $Jsubset [a,b]$ and a constant $m>0$ such that
$f(x)ge m$ for $xin J$.
The above question is from my analysis textbook and here's my attempt at proving it.
If $int_{a}^{b} f$ exists then $int_{a}^{b} f = underline{int_{a}^{b} f}$ by definition. By definition
$$underline{int_{a}^{b} f} = sup { L(P,f) mid text{P is a partition of [a,b]}}$$
Since $0$ is not an upperbound for the above set (inside the $sup$), There must be a partition $P$ such that $L(P,f) >0$. By definition,
$$L(P,f)= sum_{k=1}^{n} m_{k} (f) Delta x_k$$
If $m_k (f)$ were less than or equal to $0$ for each $k$ then $L(P,f) le 0$, thus, it must be that $m_k (f) > 0$ for some $k$. We know that $m_{k} (f) = inf { f(x) mid xin [x_{k-1} , x_{k} ] }$. Thus, we let $m=m_{k} (f)$ and it follows that $f(x)ge m$ for $xin [x_{k-1} , x_{k}]$.
Is my proof okay? Is there any easier way through this?
real-analysis proof-verification proof-writing alternative-proof riemann-integration
real-analysis proof-verification proof-writing alternative-proof riemann-integration
asked Jan 4 at 10:24
Ashish KAshish K
910613
asked Jan 4 at 10:24
Ashish KAshish K
910613
asked Jan 4 at 10:24
Ashish KAshish K
910613
asked Jan 4 at 10:24
Ashish KAshish K
910613
asked Jan 4 at 10:24
Ashish KAshish K
910613
910613
2
$begingroup$
Your proof is fine and there is no better way.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 10:29
add a comment |
2
$begingroup$
Your proof is fine and there is no better way.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 10:29
2
2
$begingroup$
Your proof is fine and there is no better way.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 10:29
$begingroup$
Your proof is fine and there is no better way.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 10:29
add a comment |
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$begingroup$
Your proof is fine and there is no better way.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 10:29