Question

Sei \(F\) eine Abbildung von \(C([a, b])\) nach \(C([a, b])\) und zwar konkret definiert durch $$ x \mapsto F x, \quad(F x)(t)=\int_{a}^{b} k(t, s) f(s, x(s)) d s $$ wobei \(k:[a, b] \times[a, b] \rightarrow \mathbb{R}\) stetig sein soll, und \(f:[a, b] \times \mathbb{R} \rightarrow \mathbb{R}\) ebenfalls stetig ist und im zweiten Argument stetig differenzierbar sein soll. Zeigen Sie unter Nutzung des Mittelwertsatzes der Differentialrechnung, dass \(F\) FRÉCHET- differenzierbar mit der FRÉCHET-Ableitung $$ \left(F^{\prime}(x) h\right)(t)=\int_{a}^{b} k(t, s) \frac{\partial f}{\partial x}(s, x(s)) h(s) d s $$ ist.

Short answer

\( F(x) \) is Fréchet differentiable with derivative \( F'(x)h = \int_{a}^{b} k(t, s) \frac{\partial f}{\partial x}(s, x(s)) h(s) \, ds \).

Step-by-step solution

Understand the Problem

We need to prove that the transformation \( F \) from \( C([a, b]) \) to \( C([a, b]) \) is Fréchet differentiable and find its derivative. The mapping is defined by \( (F x)(t) = \int_{a}^{b} k(t, s) f(s, x(s)) \, ds \), where \( k \) and \( f \) are continuous functions, and \( f \) is continuously differentiable with respect to its second argument.

Recall Fréchet Derivative Definition

The Fréchet derivative of a function \( F \) at a point \( x \) in a Banach space is a bounded linear operator \( F'(x) \), such that for \( h \) in the space, \( F(x+h)-F(x)-F'(x)h=o(\|h\|) \). This essentially means the error \( F(x+h) - F(x) - F'(x)h \) becomes negligibly small compared to \( \|h\| \) as \( \|h\| \to 0 \).

Apply Mean Value Theorem for Function \( f \)

For small perturbations \( h(s) \, \in \, C([a, b]) \), using the Mean Value Theorem for functions, we have:\[ f(s, x(s) + h(s)) - f(s, x(s)) = \frac{\partial f}{\partial x} (s, \xi(s)) h(s) \] for some \( \xi(s) \) between \( x(s) \) and \( x(s) + h(s) \). This helps in approximating \( f(s, x(s) + h(s)) \) around \( x(s) \).

Differentiate the Integral

By substituting the result from the Mean Value Theorem in our integral, we recognize that:\[(F(x+h))(t) - (F(x))(t) = \int_{a}^{b} k(t, s) \left( f(s, x(s) + h(s)) - f(s, x(s)) \right) \, ds \= \int_{a}^{b} k(t, s) \frac{\partial f}{\partial x}(s, \xi(s)) h(s) \, ds.\] By continuity, \( \xi(s) \to x(s) \) as \( h(s) \to 0 \).

Establish Fréchet Derivative Expression

Rearranging the differential terms yields:\[(F(x+h))(t) - (F(x))(t) = \int_{a}^{b} k(t, s) \frac{\partial f}{\partial x}(s, x(s)) h(s) \, ds + \epsilon(h, x)\] where \( \epsilon(h, x) = \int_{a}^{b} k(t, s) ( \frac{\partial f}{\partial x}(s, \xi(s)) - \frac{\partial f}{\partial x}(s, x(s)) ) h(s) \, ds \) tends to 0 as \( \|h\| \to 0 \). Thus, \( F'(x)h = \int_{a}^{b} k(t, s) \frac{\partial f}{\partial x}(s, x(s)) h(s) \, ds \).

Verify the Fréchet Differentiability Criteria

It remains to show that \( \epsilon(h, x) = o(\|h\|) \). Since \( \frac{\partial f}{\partial x} \) is continuous, the difference \( \frac{\partial f}{\partial x}(s, \xi(s)) - \frac{\partial f}{\partial x}(s, x(s)) \to 0 \) uniformly, ensuring \( \epsilon(h,x) \) is a higher order term compared to \( \|h\| \), thus proving the Fréchet differentiability.

Key concepts

Mean Value Theorem

The Mean Value Theorem is a fundamental concept in calculus that helps us approximate changes in continuous and differentiable functions. It effectively states that for a real-valued function that is continuous over a closed interval \[a, b\] and differentiable over the open interval \(a, b\), there exists at least one point \(c\) in the interval where the function's instantaneous rate of change (i.e., derivative) equals the average rate of change over the interval. Essentially, it guarantees that if you chart the function, there's somewhere along the curve where the tangent has the same slope as the line connecting the endpoints of the interval.

This theorem is critical in differential calculus because it often provides the means to estimate the amount by which a function changes over an interval. In the context of proving Fréchet differentiability, the Mean Value Theorem helps us relate the difference in function values to derivatives, leading to an integral form that approximates the original function. This understanding is key to managing transformations in differential calculus.

Banach Space

Banach spaces are complete normed vector spaces. This means every Cauchy sequence in a Banach space converges within the space. They play an essential role in functional analysis, a branch of mathematics that deals with functions and their generalizations. In simpler terms, a Banach space provides a structured framework where you can perform limit operations and analyze convergence behaviors.

One important feature of Banach spaces is the ability to handle infinite dimensional settings, such as function spaces like \({C([a, b])}\), the space of continuous functions on a closed interval. This space is central to the original problem, as it allows analysis of integral transformations and differentiability in a comprehensive and mathematically rigorous manner. The use of Banach spaces simplifies many complex analytical problems by ensuring that limits and functions behave predictably in all cases addressed within this space.

Differential Calculus

Differential calculus is concerned primarily with studying how functions change. It focuses on the derivative, a powerful tool that describes the rate of change of a function at any given point. The derivative is fundamental in defining concepts like tangents, velocities, and accelerations, which have broad applications from physics to engineering.

In the context of the original problem, differential calculus is employed to derive the Fréchet derivative of the transformation \(F\). Here, the challenge lies not only in finding the derivative but in proving its existence as the error term tends to zero compared to perturbations in function space. The solution utilizes differential calculus techniques to show that the derivative provides a valid linear approximation of \(F\), affirming the core principle that at very small scales, functions appear linear.

Integral Transformations

Integral transformations involve converting a function into another function via integration, often smoothing or analyzing its features in a new form. They are integral to many branches of mathematics and physics because of their ability to simplify complex problems and provide new perspectives on function behavior.

In the given exercise, an integral transformation is central to defining the function \(F\) that we need to differentiate. The expression \( (F x)(t) = \int_{a}^{b} k(t, s) f(s, x(s)) \, ds \) shows how a continuous input function \(x(s)\) can be mapped into another through a kernel \(k(t, s)\), making it easier to analyze differentiability through such transformations. By substituting results from differential calculus into this integral form, the problem highlights how transformations integrate smoothly with other mathematical operations, facilitating complex analyses like establishing differentiability or solving differential equations.