Sequence Of Functions: Differentiability And Convergence

Hey guys! Today, we're diving into the fascinating world of sequences of functions defined by integrals. We'll be tackling a problem where we have a sequence ${ u_n(x) }$ defined recursively, and we want to explore its properties, specifically its differentiability and convergence. Let's break it down step by step.

Defining the Sequence

We're given that for all ${ x }$ in the interval ${ [0, 1] }$, ${ u_0(x) = 1 }$. This is our initial function. Then, for all ${ n }$ in the set of natural numbers ${ \mathbb{N} }$, the subsequent functions are defined by the recursive relation:

${ u_{n+1}(x) = 1 + \int_{0}^{x} u_n(t - t^3) dt }$

This means each function in the sequence is built upon the previous one by integrating it over the interval ${ [0, x] }$ after a transformation of the variable. Now, let's get to the fun part: proving some cool properties about this sequence.

(1) Proving Infinite Differentiability: ${ u_n }$ is of Class ${ C^{\infty} }$

The first thing we want to show is that every function in this sequence is infinitely differentiable, i.e., of class ${ C^{\infty} }$. This means that each function ${ u_n }$ has derivatives of all orders, and all these derivatives are continuous. We'll use induction to prove this.

Base Case: ${ n = 0 }$

For ${ n = 0 }$, we have ${ u_0(x) = 1 }$. This is a constant function, and constant functions are known to be infinitely differentiable. All its derivatives are zero, which are continuous. So, the base case holds.

Inductive Step

Assume that ${ u_n(x) }$ is of class ${ C^{\infty} }$ for some ${ n \geq 0 }$. We want to show that ${ u_{n+1}(x) }$ is also of class ${ C^{\infty} }$.

We know that

${ u_{n+1}(x) = 1 + \int_{0}^{x} u_n(t - t^3) dt }$

Since ${ u_n }$ is of class ${ C^{\infty} }$ (by our inductive hypothesis), the function ${ u_n(t - t^3) }$ is also of class ${ C^{\infty} }$ because ${ t - t^3 }$ is a polynomial and thus infinitely differentiable. The composition of infinitely differentiable functions is also infinitely differentiable.

Now, we need to show that the integral ${ \int_{0}^{x} u_n(t - t^3) dt }$ is of class ${ C^{\infty} }$. By the Fundamental Theorem of Calculus, differentiation under the integral sign gives us:

${ \frac{d}{dx} \int_{0}^{x} u_n(t - t^3) dt = u_n(x - x^3) }$

Since ${ u_n(x - x^3) }$ is of class ${ C^{\infty} }$, its integral with respect to ${ x }$ is also of class ${ C^{\infty} }$. Thus, ${ u_{n+1}(x) }$ is the sum of a constant (1) and an infinitely differentiable function, which means ${ u_{n+1}(x) }$ is also of class ${ C^{\infty} }$.

Conclusion

By the principle of mathematical induction, for all ${ n \in \mathbb{N} }$, ${ u_n(x) }$ is of class ${ C^{\infty} }$. This means every function in our sequence has derivatives of all orders, and these derivatives are continuous. Awesome, right?

(2) Proving the Inequality by Induction

Now, let's prove the second part: that for all ${ n \in \mathbb{N} }$ and for all ${ x \in [0, 1] }$, ${ 0 \leq u_{n+1}(x) - u_n(x) \leq \cdots }$. We'll again use induction.

Base Case: ${ n = 0 }$

For ${ n = 0 }$, we need to show that ${ 0 \leq u_1(x) - u_0(x) }$. Let's find ${ u_1(x) }$:

${ u_1(x) = 1 + \int_{0}^{x} u_0(t - t^3) dt = 1 + \int_{0}^{x} 1 dt = 1 + x }$

So, ${ u_1(x) - u_0(x) = (1 + x) - 1 = x }$. Since ${ x \in [0, 1] }$, we have ${ 0 \leq x \leq 1 }$. Thus, ${ 0 \leq u_1(x) - u_0(x) \leq x }$. The base case holds.

Inductive Step

Assume that for some ${ n \geq 0 }$ and for all ${ x \in [0, 1] }$, ${ 0 \leq u_{n+1}(x) - u_n(x) \leq \frac{x^{n+1}}{(n+1)!} }$. We want to show that

${ 0 \leq u_{n+2}(x) - u_{n+1}(x) \leq \frac{x^{n+2}}{(n+2)!} }$

We know that

${ u_{n+2}(x) = 1 + \int_{0}^{x} u_{n+1}(t - t^3) dt }$

${ u_{n+1}(x) = 1 + \int_{0}^{x} u_n(t - t^3) dt }$

So,

${ u_{n+2}(x) - u_{n+1}(x) = \int_{0}^{x} [u_{n+1}(t - t^3) - u_n(t - t^3)] dt }$

By our inductive hypothesis,

${ 0 \leq u_{n+1}(t - t^3) - u_n(t - t^3) \leq \frac{(t - t^3)^{n+1}}{(n+1)!} }$

Since ${ 0 \leq t \leq 1 }$, we have ${ 0 \leq t - t^3 \leq t }$. Thus,

${ 0 \leq \int_{0}^{x} [u_{n+1}(t - t^3) - u_n(t - t^3)] dt \leq \int_{0}^{x} \frac{(t - t^3)^{n+1}}{(n+1)!} dt \leq \int_{0}^{x} \frac{t^{n+1}}{(n+1)!} dt }$

Now, let's evaluate the integral:

${ \int_{0}^{x} \frac{t^{n+1}}{(n+1)!} dt = \frac{1}{(n+1)!} \int_{0}^{x} t^{n+1} dt = \frac{1}{(n+1)!} \left[ \frac{t^{n+2}}{n+2} \right]_0^x = \frac{x^{n+2}}{(n+2)!} }$

Therefore,

${ 0 \leq u_{n+2}(x) - u_{n+1}(x) \leq \frac{x^{n+2}}{(n+2)!} }$

Conclusion

By the principle of mathematical induction, for all ${ n \in \mathbb{N} }$ and for all ${ x \in [0, 1] }$, ${ 0 \leq u_{n+1}(x) - u_n(x) \leq \frac{x^{n+1}}{(n+1)!} }$.

Final Thoughts

So, we've shown that the sequence of functions ${ u_n(x) }$ is infinitely differentiable and that the difference between consecutive terms is bounded by a term that goes to zero as ${ n }$ goes to infinity. This gives us some strong hints about the convergence of the sequence. Keep exploring, and you'll uncover even more cool properties about these functions!

Keep exploring!