Existence of solution to $f(x) = 1 - [int_{0}^{x} tf(t) dt]^2$
$begingroup$
Prove the existence and uniqueness of $text{ } f(x) in C^1([0,1])$ such that $text{ }f: [0,1] rightarrow [0,1]$ where begin{equation} f(x) = 1 - (int_{0}^{x} tf(t) dt)^2 end{equation}
I was able to prove uniqueness using a standard argument of using continuity to show $f(x) - g(x) = 0$ on $x in [0,1]$ is they satisfy the above integral equation. However, I'm stuck on figuring out how to show a solution exists.
My initial assumption was that one can show that a solution exists by using Banach fixed point theorem on $C^{1}[0,1]$, but I'm not sure how to bound the metric:
Let $T(f(x)) := 1 - (int_{0}^{x} tf(t) dt)^2$, then begin{equation} ||T(f(x)) - T(g(x))||_{infty} = sup_{x in [0,1]} |(int_{0}^{x} tg(t)) dt)^2 - (int_{0}^{x} tf(t) dt )^2|end{equation} But I'm not sure on how to bound the above expression to get a contraction mapping. An observation I made is the integral is an inner product $<t,f(t)>$, but when I wanted to use Cauchy Schwarz Inequality I realized I needed to use the triangle in the sup norm, from which I couldn't prove its a contraction mapping.
Any help on how to bound the sup norm or how else to proceed would be very helpful. Thank you!
real-analysis integration fixed-point-theorems
asked Dec 29 '18 at 1:25
Story123Story123
18519
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$begingroup$
Prove the existence and uniqueness of $text{ } f(x) in C^1([0,1])$ such that $text{ }f: [0,1] rightarrow [0,1]$ where begin{equation} f(x) = 1 - (int_{0}^{x} tf(t) dt)^2 end{equation}
I was able to prove uniqueness using a standard argument of using continuity to show $f(x) - g(x) = 0$ on $x in [0,1]$ is they satisfy the above integral equation. However, I'm stuck on figuring out how to show a solution exists.
My initial assumption was that one can show that a solution exists by using Banach fixed point theorem on $C^{1}[0,1]$, but I'm not sure how to bound the metric:
Let $T(f(x)) := 1 - (int_{0}^{x} tf(t) dt)^2$, then begin{equation} ||T(f(x)) - T(g(x))||_{infty} = sup_{x in [0,1]} |(int_{0}^{x} tg(t)) dt)^2 - (int_{0}^{x} tf(t) dt )^2|end{equation} But I'm not sure on how to bound the above expression to get a contraction mapping. An observation I made is the integral is an inner product $<t,f(t)>$, but when I wanted to use Cauchy Schwarz Inequality I realized I needed to use the triangle in the sup norm, from which I couldn't prove its a contraction mapping.
Any help on how to bound the sup norm or how else to proceed would be very helpful. Thank you!
real-analysis integration fixed-point-theorems
asked Dec 29 '18 at 1:25
Story123Story123
18519
$endgroup$
add a comment |
$begingroup$
Prove the existence and uniqueness of $text{ } f(x) in C^1([0,1])$ such that $text{ }f: [0,1] rightarrow [0,1]$ where begin{equation} f(x) = 1 - (int_{0}^{x} tf(t) dt)^2 end{equation}
I was able to prove uniqueness using a standard argument of using continuity to show $f(x) - g(x) = 0$ on $x in [0,1]$ is they satisfy the above integral equation. However, I'm stuck on figuring out how to show a solution exists.
My initial assumption was that one can show that a solution exists by using Banach fixed point theorem on $C^{1}[0,1]$, but I'm not sure how to bound the metric:
Let $T(f(x)) := 1 - (int_{0}^{x} tf(t) dt)^2$, then begin{equation} ||T(f(x)) - T(g(x))||_{infty} = sup_{x in [0,1]} |(int_{0}^{x} tg(t)) dt)^2 - (int_{0}^{x} tf(t) dt )^2|end{equation} But I'm not sure on how to bound the above expression to get a contraction mapping. An observation I made is the integral is an inner product $<t,f(t)>$, but when I wanted to use Cauchy Schwarz Inequality I realized I needed to use the triangle in the sup norm, from which I couldn't prove its a contraction mapping.
Any help on how to bound the sup norm or how else to proceed would be very helpful. Thank you!
real-analysis integration fixed-point-theorems
asked Dec 29 '18 at 1:25
Story123Story123
18519
$endgroup$
Prove the existence and uniqueness of $text{ } f(x) in C^1([0,1])$ such that $text{ }f: [0,1] rightarrow [0,1]$ where begin{equation} f(x) = 1 - (int_{0}^{x} tf(t) dt)^2 end{equation}
I was able to prove uniqueness using a standard argument of using continuity to show $f(x) - g(x) = 0$ on $x in [0,1]$ is they satisfy the above integral equation. However, I'm stuck on figuring out how to show a solution exists.
My initial assumption was that one can show that a solution exists by using Banach fixed point theorem on $C^{1}[0,1]$, but I'm not sure how to bound the metric:
Let $T(f(x)) := 1 - (int_{0}^{x} tf(t) dt)^2$, then begin{equation} ||T(f(x)) - T(g(x))||_{infty} = sup_{x in [0,1]} |(int_{0}^{x} tg(t)) dt)^2 - (int_{0}^{x} tf(t) dt )^2|end{equation} But I'm not sure on how to bound the above expression to get a contraction mapping. An observation I made is the integral is an inner product $<t,f(t)>$, but when I wanted to use Cauchy Schwarz Inequality I realized I needed to use the triangle in the sup norm, from which I couldn't prove its a contraction mapping.
Any help on how to bound the sup norm or how else to proceed would be very helpful. Thank you!
real-analysis integration fixed-point-theorems
real-analysis integration fixed-point-theorems
asked Dec 29 '18 at 1:25
Story123Story123
18519
asked Dec 29 '18 at 1:25
Story123Story123
18519
asked Dec 29 '18 at 1:25
Story123Story123
18519
asked Dec 29 '18 at 1:25
Story123Story123
18519
asked Dec 29 '18 at 1:25
Story123Story123
18519
18519
add a comment |
add a comment |
1 Answer
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$begingroup$
A little algebra:
begin{align}
left(int_0^x tgdtright)^2-left(int_0^x tfdtright)^2&=left[left(int_0^x tgdtright)-left(int_0^x tfdtright)right]left[left(int_0^x tgdtright)+left(int_0^x tfdtright)right]\
&=left(int_0^x t(g-f)dtright)left(int_0^x t(g+f)dtright),
end{align}
with
$$
|int_0^xt(f+g)dt|le 2int_0^xtdtle 1,
$$
and
$$
|int_0^xt(g-f)dt|le |g-f|_inftyint_0^xtdtlefrac{1}{2}|g-f|_infty.
$$
This is enough for a contraction.
answered Dec 29 '18 at 1:51
user254433user254433
2,541712
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1 Answer
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1 Answer
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$begingroup$
A little algebra:
begin{align}
left(int_0^x tgdtright)^2-left(int_0^x tfdtright)^2&=left[left(int_0^x tgdtright)-left(int_0^x tfdtright)right]left[left(int_0^x tgdtright)+left(int_0^x tfdtright)right]\
&=left(int_0^x t(g-f)dtright)left(int_0^x t(g+f)dtright),
end{align}
with
$$
|int_0^xt(f+g)dt|le 2int_0^xtdtle 1,
$$
and
$$
|int_0^xt(g-f)dt|le |g-f|_inftyint_0^xtdtlefrac{1}{2}|g-f|_infty.
$$
This is enough for a contraction.
answered Dec 29 '18 at 1:51
user254433user254433
2,541712
$endgroup$
add a comment |
$begingroup$
A little algebra:
begin{align}
left(int_0^x tgdtright)^2-left(int_0^x tfdtright)^2&=left[left(int_0^x tgdtright)-left(int_0^x tfdtright)right]left[left(int_0^x tgdtright)+left(int_0^x tfdtright)right]\
&=left(int_0^x t(g-f)dtright)left(int_0^x t(g+f)dtright),
end{align}
with
$$
|int_0^xt(f+g)dt|le 2int_0^xtdtle 1,
$$
and
$$
|int_0^xt(g-f)dt|le |g-f|_inftyint_0^xtdtlefrac{1}{2}|g-f|_infty.
$$
This is enough for a contraction.
answered Dec 29 '18 at 1:51
user254433user254433
2,541712
$endgroup$
add a comment |
$begingroup$
A little algebra:
begin{align}
left(int_0^x tgdtright)^2-left(int_0^x tfdtright)^2&=left[left(int_0^x tgdtright)-left(int_0^x tfdtright)right]left[left(int_0^x tgdtright)+left(int_0^x tfdtright)right]\
&=left(int_0^x t(g-f)dtright)left(int_0^x t(g+f)dtright),
end{align}
with
$$
|int_0^xt(f+g)dt|le 2int_0^xtdtle 1,
$$
and
$$
|int_0^xt(g-f)dt|le |g-f|_inftyint_0^xtdtlefrac{1}{2}|g-f|_infty.
$$
This is enough for a contraction.
answered Dec 29 '18 at 1:51
user254433user254433
2,541712
$endgroup$
A little algebra:
begin{align}
left(int_0^x tgdtright)^2-left(int_0^x tfdtright)^2&=left[left(int_0^x tgdtright)-left(int_0^x tfdtright)right]left[left(int_0^x tgdtright)+left(int_0^x tfdtright)right]\
&=left(int_0^x t(g-f)dtright)left(int_0^x t(g+f)dtright),
end{align}
with
$$
|int_0^xt(f+g)dt|le 2int_0^xtdtle 1,
$$
and
$$
|int_0^xt(g-f)dt|le |g-f|_inftyint_0^xtdtlefrac{1}{2}|g-f|_infty.
$$
This is enough for a contraction.
answered Dec 29 '18 at 1:51
user254433user254433
2,541712
answered Dec 29 '18 at 1:51
user254433user254433
2,541712
answered Dec 29 '18 at 1:51
user254433user254433
2,541712
answered Dec 29 '18 at 1:51
user254433user254433
2,541712
2,541712
add a comment |
add a comment |
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