micromass said:Ok, I'll give a sketch of the proof
First, convince yourself that
[tex] \int_0^\infty e^{-t^2}dt=\frac{\sqrt{\pi}}{2} [/tex]
You will need this several times.
So we have the following operator
[tex] H:\mathcal{B}(\mathbb{R}^+,\mathbb{R})\rightarrow \mathcal{B}(\mathbb{R}^+,\mathbb{R}): g \rightarrow H(g) [/tex]
With
[tex] H(g):\mathbb{R}^+\rightarrow \mathbb{R}: x\rightarrow 1+\int_0^x e^{-t^2}g(tx)dt [/tex]
Where [tex] \mathcal{B}(\mathbb{R}^+,\mathbb{R})[/tex] 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 [tex] \mathcal{B}(\mathbb{R}^+,\mathbb{R})[/tex]). We need the functions to be bounded, in order to put following metric on [tex] \mathcal{B}(\mathbb{R}^+,\mathbb{R})[/tex]:
[tex] d_\infty(f,g)=\sup_{x\in \mathbb{R}^+}{d(f(x),g(x))} [/tex]
So this makes [tex] \mathcal{B}(\mathbb{R}^+,\mathbb{R})[/tex] 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
[tex] d_\infty(H(f),H(g))\leq k d_\infty(f,g) [/tex]
So, applying the Banach fixed point theorem, yields the existence of a function f in B(R+,R)
[tex] 1, H(1), H(H(1)), H(H(H(1))),... \rightarrow f [/tex]
This function f is the unique fixed point of the operator H, and thus it satisfies
[tex] f(x)=1+\int_0^x e^{-t^2}f(tx)dt [/tex]
So, this function is indeed the one you wanted. Since f is in [tex] \mathcal{B}(\mathbb{R}^+,\mathbb{R})[/tex], we have that f is continuous and bounded.
A bounded function is one that has a finite range of values. This means that the function's output is limited to a specific range and cannot exceed certain values. In other words, the function does not have any values that go to infinity or negative infinity.
To prove that a function is bounded, you need to show that there exists a finite number M such that the absolute value of the function is always less than or equal to M. This can be done by using the definition of a bounded function and manipulating the function algebraically until you can find a specific value for M that satisfies the condition.
While a bounded function has a finite range of values, a continuous function is one that has no breaks or jumps in its graph. This means that the function can be drawn without lifting the pen from the paper. A bounded function can be discontinuous, but a continuous function is always bounded.
To prove that a function is continuous, you need to show that the limit of the function at a specific point is equal to the value of the function at that point. This can be done by using the definition of continuity and applying the limit laws to manipulate the function algebraically until the condition is satisfied.
Yes, a function can be both bounded and continuous. In fact, many common functions, such as polynomials and trigonometric functions, are both bounded and continuous. This means that the function has a finite range of values and has no breaks or jumps in its graph.