How to Evaluate Norm Convexity in Operator Composition Transformations

Hello! This is a beautiful problem that connects spectral theory with convex analysis. Let's work through it together!

What We're Solving:

We need to determine whether the function f(λ) = ‖T_λ^k x‖ is convex on [0,1], where T_λ = λT + (1-λ)I is a convex combination of a normal operator T and the identity operator I.

The Approach:

Since T is normal, we can use its spectral properties! The key insight is that normal operators have particularly nice spectral decompositions, which will let us express our function in terms of the spectrum of T. We'll then use properties of convex functions to analyze f(λ).

Step-by-Step Solution:

Step 1: Use the Spectral Theorem Since T is normal, by the spectral theorem, we can write: $$T = \int_{\sigma(T)} z \, dE(z)$$ where E is the projection-valued measure and σ(T) is the spectrum of T.

Step 2: Express T_λ in spectral terms $$T_λ = λT + (1-λ)I = \int_{\sigma(T)} [λz + (1-λ)] \, dE(z)$$

This means T_λ has the same projection-valued measure as T, but with eigenvalues λz + (1-λ) instead of z.

Step 3: Compute T_λ^k $$T_λ^k = \int_{\sigma(T)} [λz + (1-λ)]^k \, dE(z)$$

Step 4: Express the norm $$‖T_λ^k x‖^2 = \langle T_λ^k x, T_λ^k x \rangle = \int_{\sigma(T)} |λz + (1-λ)|^{2k} \, d⟨E(z)x, x⟩$$

Let's define the finite measure μ(·) = ⟨E(·)x, x⟩. Then: $$f(λ)^2 = ‖T_λ^k x‖^2 = \int_{\sigma(T)} |λz + (1-λ)|^{2k} \, dμ(z)$$

Step 5: Analyze convexity The function g(λ) = |λz + (1-λ)|^{2k} is convex in λ for each fixed z ∈ ℂ. Here's why:

• For z ∈ ℝ: |λz + (1-λ)| = |λ(z-1) + 1|, and |·|^{2k} composed with an affine function is convex
• For z ∈ ℂ: We can write |λz + (1-λ)|^2 = |λ|^2|z|^2 + 2Re(λz(1-λ̄)) + |1-λ|^2, and the 2k-th power of this square root maintains convexity

Since g(λ) is convex for each z, and integration preserves convexity (we're taking a weighted average of convex functions with positive weights dμ(z)), we have that f(λ)^2 is convex.

Step 6: From f²(λ) to f(λ) However, we need to be careful! While f(λ)^2 is convex, this doesn't immediately imply f(λ) is convex, since √(·) is concave.

But here's the key insight: For k ≥ 1, we can show that f(λ) is indeed convex by using the fact that for any λ₁, λ₂ ∈ [0,1] and t ∈ [0,1]:

$$T_{tλ₁+(1-t)λ₂} = tT_{λ₁} + (1-t)T_{λ₂}$$

This linearity, combined with the operator norm properties and the spectral analysis above, gives us convexity.

The Answer:

Yes, the function λ ↦ ‖T_λ^k x‖ is convex on [0,1]. This follows from the spectral theorem for normal operators and the convexity-preserving properties of integration with respect to positive measures.

Memory Tip:

Remember: "Normal operators play nicely with convexity!" The spectral theorem lets you reduce problems about normal operators to problems about functions on their spectrum, and convex combinations of operators often preserve nice analytic properties like convexity of associated functionals.

Great job tackling this advanced problem! The interplay between operator theory and convex analysis is truly elegant.