Proving the existence of a solution for integral equation.
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Let $u(x)$ is a continuous monotone strictly increasing function of $x$ on a finite interval $[0,b]$ and the equation $1-delta,u(x)=0$ has solution in $(0,b]$, where $delta>0$. The function $v(x)$ defined as
$$
v(x)=frac1xintlimits_0^xfracu(xi),xi1-delta,u(xi),dxi,quad xin (0,b].
$$
Prove that there is a point $x^*in (0,b]$ in which $v(x)=1$.
real-analysis integration convergence
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up vote
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down vote
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Let $u(x)$ is a continuous monotone strictly increasing function of $x$ on a finite interval $[0,b]$ and the equation $1-delta,u(x)=0$ has solution in $(0,b]$, where $delta>0$. The function $v(x)$ defined as
$$
v(x)=frac1xintlimits_0^xfracu(xi),xi1-delta,u(xi),dxi,quad xin (0,b].
$$
Prove that there is a point $x^*in (0,b]$ in which $v(x)=1$.
real-analysis integration convergence
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $u(x)$ is a continuous monotone strictly increasing function of $x$ on a finite interval $[0,b]$ and the equation $1-delta,u(x)=0$ has solution in $(0,b]$, where $delta>0$. The function $v(x)$ defined as
$$
v(x)=frac1xintlimits_0^xfracu(xi),xi1-delta,u(xi),dxi,quad xin (0,b].
$$
Prove that there is a point $x^*in (0,b]$ in which $v(x)=1$.
real-analysis integration convergence
Let $u(x)$ is a continuous monotone strictly increasing function of $x$ on a finite interval $[0,b]$ and the equation $1-delta,u(x)=0$ has solution in $(0,b]$, where $delta>0$. The function $v(x)$ defined as
$$
v(x)=frac1xintlimits_0^xfracu(xi),xi1-delta,u(xi),dxi,quad xin (0,b].
$$
Prove that there is a point $x^*in (0,b]$ in which $v(x)=1$.
real-analysis integration convergence
asked Jul 25 at 8:46
Sokuroff
263
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