Proof of an integral inequality about a decreasing continuous function
$begingroup$
$f(x)$ is a continuous function on $[0,infty)$, and it is decreasing on its domain.
Then how to prove
$int_0^x{x^2f(t),dt}ge3int_0^x{t^2f(t),dt}$ when $xge0$ ?
Maybe I can have $F(x)=int_0^x{x^2f(t),dt}-3int_0^x{t^2f(t),dt}$ and work on its derivative, but what's the derivative of $int_0^x{x^2f(t),dt}$? Can Fundamental Theorem of Calculus be used here? Or are there other ways to prove that?
integration inequality
asked Dec 31 '18 at 10:52
brokencupbrokencup
61
$endgroup$
add a comment |
$begingroup$
$f(x)$ is a continuous function on $[0,infty)$, and it is decreasing on its domain.
Then how to prove
$int_0^x{x^2f(t),dt}ge3int_0^x{t^2f(t),dt}$ when $xge0$ ?
Maybe I can have $F(x)=int_0^x{x^2f(t),dt}-3int_0^x{t^2f(t),dt}$ and work on its derivative, but what's the derivative of $int_0^x{x^2f(t),dt}$? Can Fundamental Theorem of Calculus be used here? Or are there other ways to prove that?
integration inequality
asked Dec 31 '18 at 10:52
brokencupbrokencup
61
$endgroup$
add a comment |
$begingroup$
$f(x)$ is a continuous function on $[0,infty)$, and it is decreasing on its domain.
Then how to prove
$int_0^x{x^2f(t),dt}ge3int_0^x{t^2f(t),dt}$ when $xge0$ ?
Maybe I can have $F(x)=int_0^x{x^2f(t),dt}-3int_0^x{t^2f(t),dt}$ and work on its derivative, but what's the derivative of $int_0^x{x^2f(t),dt}$? Can Fundamental Theorem of Calculus be used here? Or are there other ways to prove that?
integration inequality
asked Dec 31 '18 at 10:52
brokencupbrokencup
61
$endgroup$
$f(x)$ is a continuous function on $[0,infty)$, and it is decreasing on its domain.
Then how to prove
$int_0^x{x^2f(t),dt}ge3int_0^x{t^2f(t),dt}$ when $xge0$ ?
Maybe I can have $F(x)=int_0^x{x^2f(t),dt}-3int_0^x{t^2f(t),dt}$ and work on its derivative, but what's the derivative of $int_0^x{x^2f(t),dt}$? Can Fundamental Theorem of Calculus be used here? Or are there other ways to prove that?
integration inequality
integration inequality
asked Dec 31 '18 at 10:52
brokencupbrokencup
61
asked Dec 31 '18 at 10:52
brokencupbrokencup
61
asked Dec 31 '18 at 10:52
brokencupbrokencup
61
asked Dec 31 '18 at 10:52
brokencupbrokencup
61
asked Dec 31 '18 at 10:52
brokencupbrokencup
61
61
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your function $F$ is simply $F(x) = x^2 int_0^x f(t), dt - 3 int_0^x t^2 f(t), dt$.
Since $f$ is continuous, you can differentiate $F$ obtaining
$$
F'(x) = 2x int_0^x f(t), dt + x^2 f(x) - 3 x^2 f(x).
$$
Since $f$ is decreasing, you have that $f(t) geq f(x)$ for every $tin[0,x]$, hence
$$
int_0^x f(t), dt geq x, f(x).
$$
You then obtain $F'(x) geq 0$ for every $xgeq 0$. Since $F(0) = 0$, you can conclude that $F(x) geq 0$ for every $xgeq 0$.
answered Dec 31 '18 at 11:17
RigelRigel
11.3k11320
$endgroup$
add a comment |
$begingroup$
Edit:(This solution only works if $f$ is differentiable and $f'g+g'f$ is continuos where g(t)=(t*x^2-t^3))
Tip: Wirte the Integral as $int_0^x(x^2-3t^2)f(t)dt$ (if your inequality holds this has to be $geq 0$). Then use Partial integration. Since $f'leq 0$ the assertion follows.
answered Dec 31 '18 at 11:10
A. PA. P
1085
$endgroup$
add a comment |
$begingroup$
This inequality is about rearrangment. The following generalization is true. Assume that $g,hge 0$ are continuous and $int_0^t g(s)ds ge int_0^t h(s)ds$ for all $tle x$ and $int_0^x g(s)ds = int_0^x h(s)ds$, then it holds
$$
int_0^x g(s)f(s)dsge int_0^x h(s)f(s)ds.
$$ If $f'$ exists and is continuous, then the claim follows by integration by parts formula.
Proof: We may assume $x=1$. Since $f$ is continuous, we can approximate $f$ uniformly by a step function of the form $$
s_n(x) = sum_{j=1}^n f(frac{j}{n})1_{[frac{j-1}{n},frac{j}{n}]}(x).
$$ Let $D(t) = int_0^t (g(s)-h(s))ds$. Then,
$$
int_0^1 (g(s)-h(s))s_n(s)ds = sum_{j=1}^n f(frac{j}{n})(D(frac{j}{n})-D(frac{j-1}{n}))=sum_{j=1}^n D(frac{j}{n})(f(frac{j}{n})-f(frac{j+1}{n}))ge 0.
$$ By taking $nto infty$, we get $$
int_0^1 g(s)f(s)dsge int_0^1 h(s)f(s)ds.
$$
answered Dec 31 '18 at 11:20
SongSong
15.9k1739
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your function $F$ is simply $F(x) = x^2 int_0^x f(t), dt - 3 int_0^x t^2 f(t), dt$.
Since $f$ is continuous, you can differentiate $F$ obtaining
$$
F'(x) = 2x int_0^x f(t), dt + x^2 f(x) - 3 x^2 f(x).
$$
Since $f$ is decreasing, you have that $f(t) geq f(x)$ for every $tin[0,x]$, hence
$$
int_0^x f(t), dt geq x, f(x).
$$
You then obtain $F'(x) geq 0$ for every $xgeq 0$. Since $F(0) = 0$, you can conclude that $F(x) geq 0$ for every $xgeq 0$.
answered Dec 31 '18 at 11:17
RigelRigel
11.3k11320
$endgroup$
add a comment |
$begingroup$
Your function $F$ is simply $F(x) = x^2 int_0^x f(t), dt - 3 int_0^x t^2 f(t), dt$.
Since $f$ is continuous, you can differentiate $F$ obtaining
$$
F'(x) = 2x int_0^x f(t), dt + x^2 f(x) - 3 x^2 f(x).
$$
Since $f$ is decreasing, you have that $f(t) geq f(x)$ for every $tin[0,x]$, hence
$$
int_0^x f(t), dt geq x, f(x).
$$
You then obtain $F'(x) geq 0$ for every $xgeq 0$. Since $F(0) = 0$, you can conclude that $F(x) geq 0$ for every $xgeq 0$.
answered Dec 31 '18 at 11:17
RigelRigel
11.3k11320
$endgroup$
add a comment |
$begingroup$
Your function $F$ is simply $F(x) = x^2 int_0^x f(t), dt - 3 int_0^x t^2 f(t), dt$.
Since $f$ is continuous, you can differentiate $F$ obtaining
$$
F'(x) = 2x int_0^x f(t), dt + x^2 f(x) - 3 x^2 f(x).
$$
Since $f$ is decreasing, you have that $f(t) geq f(x)$ for every $tin[0,x]$, hence
$$
int_0^x f(t), dt geq x, f(x).
$$
You then obtain $F'(x) geq 0$ for every $xgeq 0$. Since $F(0) = 0$, you can conclude that $F(x) geq 0$ for every $xgeq 0$.
answered Dec 31 '18 at 11:17
RigelRigel
11.3k11320
$endgroup$
Your function $F$ is simply $F(x) = x^2 int_0^x f(t), dt - 3 int_0^x t^2 f(t), dt$.
Since $f$ is continuous, you can differentiate $F$ obtaining
$$
F'(x) = 2x int_0^x f(t), dt + x^2 f(x) - 3 x^2 f(x).
$$
Since $f$ is decreasing, you have that $f(t) geq f(x)$ for every $tin[0,x]$, hence
$$
int_0^x f(t), dt geq x, f(x).
$$
You then obtain $F'(x) geq 0$ for every $xgeq 0$. Since $F(0) = 0$, you can conclude that $F(x) geq 0$ for every $xgeq 0$.
answered Dec 31 '18 at 11:17
RigelRigel
11.3k11320
answered Dec 31 '18 at 11:17
RigelRigel
11.3k11320
answered Dec 31 '18 at 11:17
RigelRigel
11.3k11320
answered Dec 31 '18 at 11:17
RigelRigel
11.3k11320
11.3k11320
add a comment |
add a comment |
$begingroup$
Edit:(This solution only works if $f$ is differentiable and $f'g+g'f$ is continuos where g(t)=(t*x^2-t^3))
Tip: Wirte the Integral as $int_0^x(x^2-3t^2)f(t)dt$ (if your inequality holds this has to be $geq 0$). Then use Partial integration. Since $f'leq 0$ the assertion follows.
answered Dec 31 '18 at 11:10
A. PA. P
1085
$endgroup$
add a comment |
$begingroup$
Edit:(This solution only works if $f$ is differentiable and $f'g+g'f$ is continuos where g(t)=(t*x^2-t^3))
Tip: Wirte the Integral as $int_0^x(x^2-3t^2)f(t)dt$ (if your inequality holds this has to be $geq 0$). Then use Partial integration. Since $f'leq 0$ the assertion follows.
answered Dec 31 '18 at 11:10
A. PA. P
1085
$endgroup$
add a comment |
$begingroup$
Edit:(This solution only works if $f$ is differentiable and $f'g+g'f$ is continuos where g(t)=(t*x^2-t^3))
Tip: Wirte the Integral as $int_0^x(x^2-3t^2)f(t)dt$ (if your inequality holds this has to be $geq 0$). Then use Partial integration. Since $f'leq 0$ the assertion follows.
answered Dec 31 '18 at 11:10
A. PA. P
1085
$endgroup$
Edit:(This solution only works if $f$ is differentiable and $f'g+g'f$ is continuos where g(t)=(t*x^2-t^3))
Tip: Wirte the Integral as $int_0^x(x^2-3t^2)f(t)dt$ (if your inequality holds this has to be $geq 0$). Then use Partial integration. Since $f'leq 0$ the assertion follows.
answered Dec 31 '18 at 11:10
A. PA. P
1085
edited Dec 31 '18 at 11:34
answered Dec 31 '18 at 11:10
A. PA. P
1085
answered Dec 31 '18 at 11:10
A. PA. P
1085
answered Dec 31 '18 at 11:10
A. PA. P
1085
1085
add a comment |
add a comment |
$begingroup$
This inequality is about rearrangment. The following generalization is true. Assume that $g,hge 0$ are continuous and $int_0^t g(s)ds ge int_0^t h(s)ds$ for all $tle x$ and $int_0^x g(s)ds = int_0^x h(s)ds$, then it holds
$$
int_0^x g(s)f(s)dsge int_0^x h(s)f(s)ds.
$$ If $f'$ exists and is continuous, then the claim follows by integration by parts formula.
Proof: We may assume $x=1$. Since $f$ is continuous, we can approximate $f$ uniformly by a step function of the form $$
s_n(x) = sum_{j=1}^n f(frac{j}{n})1_{[frac{j-1}{n},frac{j}{n}]}(x).
$$ Let $D(t) = int_0^t (g(s)-h(s))ds$. Then,
$$
int_0^1 (g(s)-h(s))s_n(s)ds = sum_{j=1}^n f(frac{j}{n})(D(frac{j}{n})-D(frac{j-1}{n}))=sum_{j=1}^n D(frac{j}{n})(f(frac{j}{n})-f(frac{j+1}{n}))ge 0.
$$ By taking $nto infty$, we get $$
int_0^1 g(s)f(s)dsge int_0^1 h(s)f(s)ds.
$$
answered Dec 31 '18 at 11:20
SongSong
15.9k1739
$endgroup$
add a comment |
$begingroup$
This inequality is about rearrangment. The following generalization is true. Assume that $g,hge 0$ are continuous and $int_0^t g(s)ds ge int_0^t h(s)ds$ for all $tle x$ and $int_0^x g(s)ds = int_0^x h(s)ds$, then it holds
$$
int_0^x g(s)f(s)dsge int_0^x h(s)f(s)ds.
$$ If $f'$ exists and is continuous, then the claim follows by integration by parts formula.
Proof: We may assume $x=1$. Since $f$ is continuous, we can approximate $f$ uniformly by a step function of the form $$
s_n(x) = sum_{j=1}^n f(frac{j}{n})1_{[frac{j-1}{n},frac{j}{n}]}(x).
$$ Let $D(t) = int_0^t (g(s)-h(s))ds$. Then,
$$
int_0^1 (g(s)-h(s))s_n(s)ds = sum_{j=1}^n f(frac{j}{n})(D(frac{j}{n})-D(frac{j-1}{n}))=sum_{j=1}^n D(frac{j}{n})(f(frac{j}{n})-f(frac{j+1}{n}))ge 0.
$$ By taking $nto infty$, we get $$
int_0^1 g(s)f(s)dsge int_0^1 h(s)f(s)ds.
$$
answered Dec 31 '18 at 11:20
SongSong
15.9k1739
$endgroup$
add a comment |
$begingroup$
This inequality is about rearrangment. The following generalization is true. Assume that $g,hge 0$ are continuous and $int_0^t g(s)ds ge int_0^t h(s)ds$ for all $tle x$ and $int_0^x g(s)ds = int_0^x h(s)ds$, then it holds
$$
int_0^x g(s)f(s)dsge int_0^x h(s)f(s)ds.
$$ If $f'$ exists and is continuous, then the claim follows by integration by parts formula.
Proof: We may assume $x=1$. Since $f$ is continuous, we can approximate $f$ uniformly by a step function of the form $$
s_n(x) = sum_{j=1}^n f(frac{j}{n})1_{[frac{j-1}{n},frac{j}{n}]}(x).
$$ Let $D(t) = int_0^t (g(s)-h(s))ds$. Then,
$$
int_0^1 (g(s)-h(s))s_n(s)ds = sum_{j=1}^n f(frac{j}{n})(D(frac{j}{n})-D(frac{j-1}{n}))=sum_{j=1}^n D(frac{j}{n})(f(frac{j}{n})-f(frac{j+1}{n}))ge 0.
$$ By taking $nto infty$, we get $$
int_0^1 g(s)f(s)dsge int_0^1 h(s)f(s)ds.
$$
answered Dec 31 '18 at 11:20
SongSong
15.9k1739
$endgroup$
This inequality is about rearrangment. The following generalization is true. Assume that $g,hge 0$ are continuous and $int_0^t g(s)ds ge int_0^t h(s)ds$ for all $tle x$ and $int_0^x g(s)ds = int_0^x h(s)ds$, then it holds
$$
int_0^x g(s)f(s)dsge int_0^x h(s)f(s)ds.
$$ If $f'$ exists and is continuous, then the claim follows by integration by parts formula.
Proof: We may assume $x=1$. Since $f$ is continuous, we can approximate $f$ uniformly by a step function of the form $$
s_n(x) = sum_{j=1}^n f(frac{j}{n})1_{[frac{j-1}{n},frac{j}{n}]}(x).
$$ Let $D(t) = int_0^t (g(s)-h(s))ds$. Then,
$$
int_0^1 (g(s)-h(s))s_n(s)ds = sum_{j=1}^n f(frac{j}{n})(D(frac{j}{n})-D(frac{j-1}{n}))=sum_{j=1}^n D(frac{j}{n})(f(frac{j}{n})-f(frac{j+1}{n}))ge 0.
$$ By taking $nto infty$, we get $$
int_0^1 g(s)f(s)dsge int_0^1 h(s)f(s)ds.
$$
answered Dec 31 '18 at 11:20
SongSong
15.9k1739
edited Dec 31 '18 at 11:56
answered Dec 31 '18 at 11:20
SongSong
15.9k1739
answered Dec 31 '18 at 11:20
SongSong
15.9k1739
answered Dec 31 '18 at 11:20
SongSong
15.9k1739
15.9k1739
add a comment |
add a comment |
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