Question

Let \(X\) be a closed subspace of \(C[0,1]\) such that every element of \(X\) is a continuously differentiable function on \([0,1]\). Show that \(X\) is finitedimensional. Hint: Let \(T: X \rightarrow C[0,1]\) be defined for \(f \in X\) by \(T(f)=f^{\prime} .\) Show that the graph of \(T\) is closed: If \(f_{n} \rightarrow f\) uniformly and \(f_{n}^{\prime} \rightarrow g\) uniformly, then \(f^{\prime}=g .\) Therefore \(T\) is continuous by the closed graph theorem. Thus for some \(n \in \mathbf{N}\) we have \(\left\|f^{\prime}\right\|_{\infty} \leq n\) whenever \(f \in X\) satisfies \(\|f\|_{\infty} \leq 1\). Let \(x_{i}=\frac{i}{4 n}\) for \(i=0,1, \ldots, 4 n\). Define an operator \(S: X \rightarrow\) \(\mathbf{R}^{4 n+1}\) by \(S(f)=\left\\{f\left(x_{i}\right)\right\\} .\) We claim that \(S\) is one-to-one. It is enough to show that if \(\|f\|_{\infty}=1\), then for some \(i, S(f)\left(x_{i}\right) \neq 0 .\) Assume that this is not true. If \(f(x)=1\) and \(x \in\left(\frac{i}{4 n}, \frac{i+1}{4 n}\right)\), then by the Lagrange mean value theorem we have \(\left|f(x)-f\left(\frac{i}{4 n}\right)\right|=\left|f^{\prime}(\xi)\right|\left|x-\frac{i}{4 n}\right| \leq n \cdot \frac{1}{4 n}\), a contradiction. Therefore \(\operatorname{dim}(X) \leq 4 n+1\)

Short answer

The space \(X\) is finite-dimensional with \(\operatorname{dim}(X) \leq 4n+1\).

Step-by-step solution

Define the Operator T

Define the operator \(T: X \rightarrow C[0,1]\) such that for \(f \in X\), \(T(f) = f'\). This means \(T\) takes a continuously differentiable function in \(X\) and maps it to its derivative in \(C[0,1]\).

Establish Closed Graph of T

Show that the graph of \(T\) is closed. This involves demonstrating that if \(f_n \rightarrow f\) uniformly and \(f_n' \rightarrow g\) uniformly, then \(f' = g\). Use uniform convergence properties to complete this step.

Apply the Closed Graph Theorem

By the closed graph theorem, since the graph of \(T\) is closed, \(T\) must be continuous. This continuity implies that for some \(n \in \mathbb{N}\), \(\lVert f' \rVert_{\infty} \leq n\) whenever \(f \in X\) satisfies \(\lVert f \rVert_{\infty} \leq 1\).

Define Points and Operator S

Let \(x_i = \frac{i}{4n}\) for \(i = 0, 1, \ldots, 4n\). Define an operator \(S: X \rightarrow \mathbb{R}^{4n+1}\) by \(S(f) = \{ f(x_i) \}\). This operator maps a function to its values at these specified points.

Show That S is One-to-One

To prove \(S\) is one-to-one, it suffices to show that if \(\lVert f \rVert_{\infty} = 1\), then for some \(i\), \(S(f)(x_i) \eq 0\). Assume the contrary and apply the Lagrange mean value theorem. If \(f(x)=1\) and \(x \in (\frac{i}{4n}, \frac{i+1}{4n})\), then \(\lvert f(x) - f(\frac{i}{4n}) \rvert = \lvert f'(\xi) \rvert \cdot \lvert x - \frac{i}{4n} \rvert \leq n \cdot \frac{1}{4n}\). This leads to a contradiction.

Conclude Dimensionality of X

Since \(S\) is one-to-one, and \(S\) maps into \(\mathbb{R}^{4n+1}\), the dimension of \(X\) must be at most \(4n+1\). Thus, \(\operatorname{dim}(X) \leq 4n+1\).

Key concepts

closed graph theorem

The closed graph theorem is a fundamental result in functional analysis, which states that if a linear operator between two Banach spaces has a closed graph, then the operator is continuous. A graph of an operator \(T\), denoted \( \text{graph}(T) \), consists of pairs \((f, T(f))\), where \(f\) is in the domain of \(T\). To show that the graph is closed, we must demonstrate that if a sequence \( f_n \to f \) uniformly and \( T(f_n) \to g \) uniformly, then \( T(f) = g \). This means if the function and its image under \( T \) both converge to some limits, then those limits must satisfy the operator's definition. This theorem is used to prove the continuity of \(T\) in the problem, implying that for every continuously differentiable function \( f \) in the subspace \( X \), its derivative remains bounded. Thus, the closed graph theorem helps establish that our space \( X \) is finite-dimensional.

• Verify linearity of the operator \( T(f) = f' \).
• Show the uniform convergence properties required.
• Apply the closed graph theorem to infer continuity.

uniform convergence

Uniform convergence is crucial for analyzing the behavior of sequences of functions. A sequence of functions \( f_n \) converges uniformly to a function \( f \) on a set \( D \) if, for every \( \epsilon > 0 \), there exists an \( N \) such that for all \( n > N \) and all \( x \) in \( D \), the inequality \( | f_n(x) - f(x) | < \epsilon \) holds. This ensures that the functions \( f_n \) approach \( f \) uniformly, not varying with \( x \). In our problem, both \( f_n \to f \) and \( f_n' \to g \) uniformly ensure that the pointwise limit of the derivatives equals the derivative of the limit function. This condition is used to validate the closed graph of the operator \( T \).

• Define the notion of uniform convergence.
• Compare it with pointwise convergence.
• Establish the implications of uniform convergence for our functions and their derivatives.

one-to-one operator

An operator \( S \) is one-to-one (injective) if different inputs are mapped to different outputs. In mathematical terms, \( S(f) = S(g) \) implies \( f = g \). To show that the operator \( S \) in our problem is one-to-one, we need to prove that no two different functions in \( X \) map to the same point in \( \mathbb{R}^{4n+1} \). This is done by considering a function \( f \) such that \( \| f \|_ \infty = 1 \), and assuming that it does not take zero at any finite point \( x_i \). Using the Lagrange mean value theorem, we derive a contradiction, proving that such an \( f \) must necessarily be zero somewhere, establishing the one-to-one nature of \( S \).

• Define one-to-one operators.
• Apply the Lagrange mean value theorem for proof.
• Show why failing to meet this property leads to a contradiction.