Let \(V = C(\mathbb{R})\) be the vector space of all continuous real-valued functions on \(\mathbb{R}\). Let \(T: V \to V\) be the linear operator defined by: \[(Tf)(x) = \int_0^x f(t) \, dt.\] Prove that \(T\) has no characteristic values (eigenvalues).
Solution
Goal: Prove that there does not exist any non-zero function \(f \in C(\mathbb{R})\) and scalar \(\lambda \in \mathbb{R}\) such that \(Tf = \lambda f\).
1. Setting up the Eigenvalue Equation:Proof:
1.1. Suppose, for contradiction, that \(\lambda \in \mathbb{R}\) is an eigenvalue of \(T\) with non-zero eigenfunction \(f \in C(\mathbb{R}) \setminus \{0\}\).
1.2. The eigenvalue equation is:\[(Tf)(x) = \int_0^x f(t) \, dt = \lambda f(x) \quad \text{for all } x \in \mathbb{R}.\]
2. Case 1: \(\lambda = 0\):Proof:
2.1. If \(\lambda = 0\), then:\[\int_0^x f(t) \, dt = 0 \quad \text{for all } x \in \mathbb{R}.\]
2.2. By the Fundamental Theorem of Calculus, since \(f\) is continuous, the function \(F(x) = \int_0^x f(t) \, dt\) is differentiable and \(F'(x) = f(x)\).
2.3. Differentiating both sides with respect to \(x\):\[f(x) = \frac{d}{dx}(0) = 0 \quad \text{for all } x \in \mathbb{R}.\]
2.4. Thus \(f\) is the zero function, which contradicts \(f \ne 0\).
2.5. Therefore, \(\lambda = 0\) is not an eigenvalue.
3. Case 2: \(\lambda \ne 0\):Proof:
3.1. If \(\lambda \ne 0\), then:\[f(x) = \frac{1}{\lambda} \int_0^x f(t) \, dt.\]
3.2. Initial Condition at \(x = 0\):Evaluating at \(x = 0\):\[f(0) = \frac{1}{\lambda} \int_0^0 f(t) \, dt = 0.\]
3.3. Differentiability and Differential Equation:Since \(f\) is continuous, the integral \(\int_0^x f(t) \, dt\) is \(C^1\), which implies \(f\) is differentiable.Differentiating both sides with respect to \(x\):\[f'(x) = \frac{1}{\lambda} f(x) \iff f'(x) - \frac{1}{\lambda} f(x) = 0.\]
3.4. This is a first-order linear ordinary differential equation.
3.5. Multiplying by the integrating factor \(e^{-x/\lambda}\):\[\frac{d}{dx} \left( e^{-x/\lambda} f(x) \right) = 0 \implies f(x) = C e^{x/\lambda} \quad \text{for some constant } C \in \mathbb{R}.\]
3.6. Using the initial condition \(f(0) = 0\):\[f(0) = C e^0 = C = 0.\]
3.7. Thus \(f(x) = 0 \cdot e^{x/\lambda} = 0\) for all \(x \in \mathbb{R}\).
3.8. This again contradicts \(f \ne 0\).
4. Conclusion:\(T\) has no characteristic values in \(\mathbb{R}\). Q.E.D.