Problem: Let H be a separable Hilbert space, and let T: H→ H be a bounded self-adjoint operator. Consider the following questions: 1. Spectral Theorem (Compact Case): Suppose T is a compact self-adjoint operator. By the spectral theorem, there exists a sequence of real eigenvalues {}, and an orthonormal basis {e} of H, consisting of eigenvectors of T. Prove the spectral decomposition of T, i.e., show that: Tx=(x, en)en for all a € H, n=1 where 0 as n → ∞. → 2. Bounded Linear Operator and Resolvent Set: Let T be a bounded self-adjoint operator on H. Define the resolvent set p(T) of T as the set of AEC such that the operator (T - XI)-1 exists and is bounded. Show that the resolvent set p(T) is non-empty and that the spectrum σ(T) of T lies in R. 3. Spectral Radius Formula: For a bounded linear operator T on a Banach space, the spectral radius r(T) is defined as r(T) = sup{|A| A € σ(T)}. : Prove that the spectral radius of T can be computed as r(T) = lim ||T||1/n x+u 4. Compact Operator and Eigenvalue Accumulation: Suppose T is a compact operator on H with non-zero eigenvalues {n}. Prove that 0 is the only possible accumulation point of the eigenvalues of T. Hints and Guidance: 1. For the first part, start by applying the spectral theorem for compact self-adjoint operators, and show how the operator T can be expressed using its eigenvalues and eigenvectors. 2. For the second part, use properties of self-adjoint operators to show that the spectrum is real, and apply the fact that T - AI is invertible for some complex A. 3. For the spectral radius formula, apply Gelfand's formula for the norm of powers of the operator, making use of the properties of bounded operators. 4. In the fourth part, use the fact that compact operators on infinite-dimensional spaces can only have 0 as an accumulation point for their eigenvalues.