Question

Show that the mapping \(T: C[0,1] \rightarrow C[0,1]\) defined by $$ (T f)(s)=\int_{0}^{s} f, \quad s \in[0,1], \quad f \in C[0,1] $$ is linear and continuous.

Short answer

To show that the mapping \(T: C[0,1] \rightarrow C[0,1]\) defined by \( (T f)(s)=\int_{0}^{s} f, \) is linear and continuous, we followed a two-step procedure. In Step 1, we proved linearity by showing that two properties hold for all functions in \(C[0,1]\) and all scalars in \(\mathbb{R}\). In Step 2, we proved continuity by showing that the function preserves convergence for sequences of continuous functions. Thus, the given mapping is proven to be both linear and continuous.

Step-by-step solution

Prove Linearity

To prove that \(T\) is linear, we need to show that the following two properties hold for all \(f, g \in C[0, 1]\) and all scalars \(a, b \in \mathbb{R}\): 1. \(T(af + bg) = aT(f) + bT(g)\) 2. \(T(f + g) = T(f) + T(g)\) Let's start by proving the first property: $$ (T(af + bg))(s) = \int_{0}^{s} (af + bg)(t) dt = a\int_{0}^{s} f(t) dt + b\int_{0}^{s} g(t) dt = a(T f)(s) + b(T g)(s) $$ For the second property: $$ (T(f + g))(s) = \int_{0}^{s} (f + g)(t) dt = \int_{0}^{s} f(t) dt + \int_{0}^{s} g(t) dt = (T f)(s) + (T g)(s) $$ Both properties hold, and therefore, \(T\) is linear.

Prove Continuity

To prove continuity, we need to show that the function preserves convergence, i.e., if a sequence of continuous functions \(\{f_n\}_{n\in\mathbb{N}}\) converges uniformly to a continuous function \(f\) on \([0,1]\), then the sequence of transformed functions \(\{T f_n\}_{n\in\mathbb{N}}\) converges uniformly to \(T f\) on \([0,1]\). Let \(\{f_n\}_{n\in\mathbb{N}}\) be a sequence of continuous functions on \([0,1]\) converging uniformly to a continuous function \(f\). By the Fundamental Theorem of Calculus, for each \(n\in\mathbb{N}\), the function \((T f_n)(s)=\int_{0}^{s} f_n(t) dt\) is differentiable on \((0,1)\), and its derivative \((T f_n)'(s)=f_n(s)\). Since the sequence \(\{f_n\}_{n\in\mathbb{N}}\) converges uniformly to \(f\), for any \(\epsilon>0\), there exists \(N\in\mathbb{N}\) such that for all \(n\geq N\) and all \(s\in[0,1]\), we have \(|f_n(s)-f(s)|<\frac{\epsilon}{2}\). Using the Mean Value Theorem, for each \(n\in\mathbb{N}\) and for each \(s\in(0,1)\), there exists \(c_s\in(0,s)\) such that $$ |(T f_n)(s)-(T f)(s)|=\left|\int_{0}^{s} [f_n(t)-f(t)] dt\right|=|f_n(c_s)-f(c_s)||s-0|<\frac{\epsilon}{2}s. $$ Taking the supremum over \(s\in[0,1]\) gives $$ \| T f_n - T f\|_\infty = \sup_{s\in [0,1]} |(T f_n)(s)-(T f)(s)|<\frac{\epsilon}{2} \leq \epsilon, $$ for all \(n\geq N\). Thus, the sequence \(\{T f_n\}_{n\in\mathbb{N}}\) converges uniformly to \(T f\) on \([0,1]\). This shows that \(T\) is continuous. In conclusion, the mapping \(T: C[0,1] \rightarrow C[0,1]\) defined by $$ (T f)(s)=\int_{0}^{s} f, \quad s \in[0,1], \quad f \in C[0,1] $$ is linear and continuous.

Key concepts

Continuity of Functions

Continuity is a fundamental concept in calculus and mathematical analysis. A function is continuous if small changes in the input result in small changes in the output. An intuitive way to understand this is that you can draw the graph of a continuous function without lifting your pen from the paper.

Consider the function \( T: C[0,1] \rightarrow C[0,1] \) given by \( (T f)(s) = \int_{0}^{s} f(t) \, dt \). For \( T \) to be continuous, it must preserve the uniform convergence of function sequences. This means if a sequence of functions \( \{f_n\} \) converging to \( f \) uniformly, then \( \{T f_n\} \) also converges to \( T f \) uniformly.

Uniform convergence is crucial in this context because it ensures that the limit function \( T f \) inherits the properties, like continuity, from the sequence. We've shown that the mapping \( T \) preserves such convergence, ensuring that it is continuous.

Uniform Convergence

Uniform convergence is a stronger form of convergence for function sequences than pointwise convergence. It ensures that the functions in the sequence converge to a limit function at the same rate across the entire interval. This property is essential in maintaining the continuous nature of function transformations.

For our context with the operator \( T \), if a sequence \( \{f_n\} \) converges uniformly to \( f \), then \( \{T f_n\} = \left\{ \int_{0}^{s} f_n(t) \, dt \right\} \) converges uniformly to \( T f = \int_{0}^{s} f(t) \, dt \). Uniform convergence is pivotal because it allows swapping limits and integrals, retaining properties like boundedness and continuity from \( \{f_n\} \) to \( T f \).

This characteristic of \( T \) leveraging uniform convergence helps ensure results meaningful and robust, making it a linchpin concept in guaranteeing continuity and further illustrating the linearity of the transformation.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, two core operations in calculus. It states that if a function is continuous on an interval, the integral of its derivative over that interval recovers the original function, up to a constant.The exercise uses this theorem to illustrate why each \( T f_n \) remains continuous even after transformation by the integral operator \( T \). Specifically, since the original functions \( f_n \) are continuous, their integrals \( \int_{0}^{s} f_n(t) \, dt \) are guaranteed to remain differentiable and thus continuous by this theorem.

By leveraging the Fundamental Theorem of Calculus, we ensure that with each transformation \( T f_n \), the continuity of the limit function \( T f \) is preserved across the domain \([0,1]\). This concrete demonstration underscores the seamless link between continuous function properties and their integral transformations.

Mean Value Theorem

The Mean Value Theorem (MVT) is a crucial concept in calculus, explaining that for a differentiable function on a closed interval, there exists a point in the open interval where the derivative of the function equals the average rate of change over the interval.

In our solution, the MVT helps us understand uniform convergence further. When considering \( T f_n \), the theorem confirms that at some point \( c_s \), the difference between \( T f_n(s) \) and \( T f(s) \) can be controlled and made arbitrarily small as in \(|f_n(c_s) - f(c_s)|\). This ensures that the impact of the transformation diminishes with respect to \( \epsilon \), helping us conclude uniform convergence back to \( T f \).

Within this context, the MVT provides the rigor needed to guarantee transformation properties align with their expected theoretical outcomes, yielding uniformity and continuity that support our proof of \( T \)'s continuity.