Functional equation with integral inequality
$begingroup$
Given the function $f:[0,infty)toBbb R$.
a) $f$ is an increasing function
b) $F(0)=0$ and $F(x+y)le F(x)+F(y)$ for all $x,y$ in the domain. ($F$ is the primitive of $f$)
Find $f$. Can somebody help me with some ideas, please?
calculus integration functions
edited Dec 19 '18 at 20:30
jayant98
657318
asked Dec 19 '18 at 19:16
GaboruGaboru
1677
$endgroup$
add a comment |
$begingroup$
Given the function $f:[0,infty)toBbb R$.
a) $f$ is an increasing function
b) $F(0)=0$ and $F(x+y)le F(x)+F(y)$ for all $x,y$ in the domain. ($F$ is the primitive of $f$)
Find $f$. Can somebody help me with some ideas, please?
calculus integration functions
edited Dec 19 '18 at 20:30
jayant98
657318
asked Dec 19 '18 at 19:16
GaboruGaboru
1677
$endgroup$
add a comment |
$begingroup$
Given the function $f:[0,infty)toBbb R$.
a) $f$ is an increasing function
b) $F(0)=0$ and $F(x+y)le F(x)+F(y)$ for all $x,y$ in the domain. ($F$ is the primitive of $f$)
Find $f$. Can somebody help me with some ideas, please?
calculus integration functions
edited Dec 19 '18 at 20:30
jayant98
657318
asked Dec 19 '18 at 19:16
GaboruGaboru
1677
$endgroup$
Given the function $f:[0,infty)toBbb R$.
a) $f$ is an increasing function
b) $F(0)=0$ and $F(x+y)le F(x)+F(y)$ for all $x,y$ in the domain. ($F$ is the primitive of $f$)
Find $f$. Can somebody help me with some ideas, please?
calculus integration functions
calculus integration functions
edited Dec 19 '18 at 20:30
jayant98
657318
asked Dec 19 '18 at 19:16
GaboruGaboru
1677
edited Dec 19 '18 at 20:30
jayant98
657318
asked Dec 19 '18 at 19:16
GaboruGaboru
1677
edited Dec 19 '18 at 20:30
jayant98
657318
edited Dec 19 '18 at 20:30
jayant98
657318
edited Dec 19 '18 at 20:30
jayant98
657318
657318
asked Dec 19 '18 at 19:16
GaboruGaboru
1677
asked Dec 19 '18 at 19:16
GaboruGaboru
1677
asked Dec 19 '18 at 19:16
GaboruGaboru
1677
1677
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
hint
Let $$F(x)=int_0^xf(t)dt$$
then for $xge yge 0$,
$$F(x+y)-F(y)=int_y^{x+y}f(t)dt$$
$$=int_0^xf(t+y)dt$$
thus
$$F(x+y)-F(y)-F(x)=int_0^x(f(t+y)-f(t))dtge 0$$
answered Dec 19 '18 at 20:20
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
hint
Let $$F(x)=int_0^xf(t)dt$$
then for $xge yge 0$,
$$F(x+y)-F(y)=int_y^{x+y}f(t)dt$$
$$=int_0^xf(t+y)dt$$
thus
$$F(x+y)-F(y)-F(x)=int_0^x(f(t+y)-f(t))dtge 0$$
answered Dec 19 '18 at 20:20
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
hint
Let $$F(x)=int_0^xf(t)dt$$
then for $xge yge 0$,
$$F(x+y)-F(y)=int_y^{x+y}f(t)dt$$
$$=int_0^xf(t+y)dt$$
thus
$$F(x+y)-F(y)-F(x)=int_0^x(f(t+y)-f(t))dtge 0$$
answered Dec 19 '18 at 20:20
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
hint
Let $$F(x)=int_0^xf(t)dt$$
then for $xge yge 0$,
$$F(x+y)-F(y)=int_y^{x+y}f(t)dt$$
$$=int_0^xf(t+y)dt$$
thus
$$F(x+y)-F(y)-F(x)=int_0^x(f(t+y)-f(t))dtge 0$$
answered Dec 19 '18 at 20:20
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
hint
Let $$F(x)=int_0^xf(t)dt$$
then for $xge yge 0$,
$$F(x+y)-F(y)=int_y^{x+y}f(t)dt$$
$$=int_0^xf(t+y)dt$$
thus
$$F(x+y)-F(y)-F(x)=int_0^x(f(t+y)-f(t))dtge 0$$
answered Dec 19 '18 at 20:20
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 19 '18 at 20:20
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 19 '18 at 20:20
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 19 '18 at 20:20
hamam_Abdallahhamam_Abdallah
38k21634
38k21634
add a comment |
add a comment |
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