Prove Bounded and Continuous Function

[ccrraappy]

Homework Statement

Prove that the following function is continuous and bounded in R+

Homework Equations

f(x) = 1 + \int_0^x e^{-t^2} f(xt) dt \qquad \forall x \geq 0

The Attempt at a Solution

I thought of using Taylor Formula, but the integral is t instead of x, so now i have no idea what to do. Help please?

 

[micromass]

Hmm, at first it doesn't even is clear why such a function f exists...

So at first, I would define the following function

H(g):\mathbb{R}^+\rightarrow \mathbb{R}:x\rightarrow 1+\int_{0}^x e^{-t^2}g(tx)dt

Then H would be an operator

H: \mathcal{C}(\mathbb{R}^+,\mathbb{R})\rightarrow \mathcal{C}(\mathbb{R}^+,\mathbb{R})

The function f would then be the fixed point of H. In fact we can (probably) prove that the sequence 1, H(1), H(H(1)), H(H(H(1))), H(H(H(H(1)))), ... converges uniformly to this fixed point f. The only thing we need to do now is show that at every step, the function is bounded and continuous. I'll leave that up to you, say something if you need more help...

 

[ccrraappy]

I think you certainly got the part about using fixed point right, because we just learned 3 fixed point theorems tarski, banach and brouwer. Someone told me to try contraction on the function but, sorry, i am all lost. Can you tell me more about it? Like giving some steps but leaving out some intermediate proofs.

 

[micromass]

Ok, I'll give a sketch of the proof

First, convince yourself that

\int_0^\infty e^{-t^2}dt=\frac{\sqrt{\pi}}{2}

You will need this several times.

So we have the following operator

H:\mathcal{B}(\mathbb{R}^+,\mathbb{R})\rightarrow \mathcal{B}(\mathbb{R}^+,\mathbb{R}): g \rightarrow H(g)

With

H(g):\mathbb{R}^+\rightarrow \mathbb{R}: x\rightarrow 1+\int_0^x e^{-t^2}g(tx)dt

Where \mathcal{B}(\mathbb{R}^+,\mathbb{R}) is the set of all BOUNDED continuous functions of R+ to R. (You need to check that H is well defined, i.e. that H(g) is bounded and continuous if g is in \mathcal{B}(\mathbb{R}^+,\mathbb{R})). We need the functions to be bounded, in order to put following metric on \mathcal{B}(\mathbb{R}^+,\mathbb{R}):

d_\infty(f,g)=\sup_{x\in \mathbb{R}^+}{d(f(x),g(x))}

So this makes \mathcal{B}(\mathbb{R}^+,\mathbb{R}) a metric space. In order to apply the fixed point theorem of Banach, we need to know that this space is complete (this is easy to check, although the answer will probably be in your course). We also need to show that H is a contraction, i.e. there exists a k<1

d_\infty(H(f),H(g))\leq k d_\infty(f,g)

So, applying the Banach fixed point theorem, yields the existence of a function f in B(R+,R)

1, H(1), H(H(1)), H(H(H(1))),... \rightarrow f

This function f is the unique fixed point of the operator H, and thus it satisfies

f(x)=1+\int_0^x e^{-t^2}f(tx)dt

So, this function is indeed the one you wanted. Since f is in \mathcal{B}(\mathbb{R}^+,\mathbb{R}), we have that f is continuous and bounded.

 

[ccrraappy]

Thanks, I will take some time to digest this.

Ok, I'll give a sketch of the proof

First, convince yourself that

\int_0^\infty e^{-t^2}dt=\frac{\sqrt{\pi}}{2}

You will need this several times.

So we have the following operator

H:\mathcal{B}(\mathbb{R}^+,\mathbb{R})\rightarrow \mathcal{B}(\mathbb{R}^+,\mathbb{R}): g \rightarrow H(g)

With

H(g):\mathbb{R}^+\rightarrow \mathbb{R}: x\rightarrow 1+\int_0^x e^{-t^2}g(tx)dt

Where \mathcal{B}(\mathbb{R}^+,\mathbb{R}) is the set of all BOUNDED continuous functions of R+ to R. (You need to check that H is well defined, i.e. that H(g) is bounded and continuous if g is in \mathcal{B}(\mathbb{R}^+,\mathbb{R})). We need the functions to be bounded, in order to put following metric on \mathcal{B}(\mathbb{R}^+,\mathbb{R}):

d_\infty(f,g)=\sup_{x\in \mathbb{R}^+}{d(f(x),g(x))}

So this makes \mathcal{B}(\mathbb{R}^+,\mathbb{R}) a metric space. In order to apply the fixed point theorem of Banach, we need to know that this space is complete (this is easy to check, although the answer will probably be in your course). We also need to show that H is a contraction, i.e. there exists a k<1

d_\infty(H(f),H(g))\leq k d_\infty(f,g)

So, applying the Banach fixed point theorem, yields the existence of a function f in B(R+,R)

1, H(1), H(H(1)), H(H(H(1))),... \rightarrow f

This function f is the unique fixed point of the operator H, and thus it satisfies

f(x)=1+\int_0^x e^{-t^2}f(tx)dt

So, this function is indeed the one you wanted. Since f is in \mathcal{B}(\mathbb{R}^+,\mathbb{R}), we have that f is continuous and bounded.