Functional Analysis 14 | Example Operator Norm [dark version]

TL;DR

Define T_G on C([0,1]) by T_G(f)=∫_0^1 G(x)f(x)dx and measure inputs with the supremum norm.

Briefing Cornell Notes

Briefing

A linear functional built from a fixed continuous, zero-free function G on the unit interval has an operator norm equal exactly to the integral of |G| over [0,1]. That result matters because it turns an abstract “supremum over all inputs” definition into a concrete number you can compute directly—an essential move in functional analysis when checking boundedness and finding sharp norms.

The setup uses X = C([0,1]) with the supremum norm ||f||∞ = sup_{x∈[0,1]} |f(x)|, and Y = ℝ or ℂ with the absolute value norm. For a chosen continuous function G with no zeros, define an operator T (indexed by G) that maps f ∈ X to the scalar

T_G(f) = ∫_0^1 G(x) f(x) dx.

The central questions are whether T_G is bounded and what its operator norm ||T_G|| is. By definition,

||T_G|| = sup_{f≠0} |T_G(f)| / ||f||∞.

A standard scaling trick reduces the search to functions with ||f||∞ = 1, so the norm becomes

||T_G|| = sup_{||f||∞=1} |∫_0^1 G(x) f(x) dx|.

To prove boundedness and get an upper bound, the absolute value is moved inside the integral: |∫ Gf| ≤ ∫ |G||f|. Since |f(x)| ≤ ||f||∞ = 1 for all x, the integrand satisfies |G(x)||f(x)| ≤ |G(x)|. This yields

|T_G(f)| ≤ ∫_0^1 |G(x)| dx,

so ||T_G|| cannot exceed ∫_0^1 |G(x)| dx. The integral is finite because |G| is continuous and nonnegative on a compact interval.

The remaining step is to show the upper bound is actually attained. The method is to construct a specific input function that “matches” the phase of G. Define

H(x) = \overline{G(x)}/|G(x)|,

which is well-defined because G has no zeros. Then ||H||∞ = 1. Plugging H into T_G gives

T_G(H) = ∫_0^1 G(x)·(\overline{G(x)}/|G(x)|) dx.

Taking absolute values and using that the integrand becomes |G(x)| (the complex conjugation ensures the product produces |G|^2), the absolute value of T_G(H) equals ∫_0^1 |G(x)| dx. This provides a matching lower bound, forcing equality:

||T_G|| = ∫_0^1 |G(x)| dx.

In short: the operator norm of the integral functional f ↦ ∫ Gf is exactly the L^1-size of |G| on [0,1], and the proof hinges on a sharp choice of test function H that saturates the supremum.

Cornell Notes

On X = C([0,1]) with the supremum norm and Y = ℝ or ℂ with absolute value, the operator T_G defined by T_G(f) = ∫_0^1 G(x)f(x)dx is bounded. Using |∫ Gf| ≤ ∫ |G||f| and the constraint ||f||∞ = 1, one gets the upper bound ||T_G|| ≤ ∫_0^1 |G(x)|dx. To show this is sharp, choose H(x) = \overline{G(x)}/|G(x)|, which is valid because G has no zeros; then ||H||∞ = 1. Substituting H into T_G makes |T_G(H)| equal to ∫_0^1 |G(x)|dx, so the supremum is attained and ||T_G|| equals that integral.

Why can the supremum in the operator norm be restricted to functions with ||f||∞ = 1?

Because the operator norm is defined as sup_{f≠0} |T_G(f)|/||f||∞. For any nonzero f, scaling by α = 1/||f||∞ produces a new function g = f/||f||∞ with ||g||∞ = 1 and T_G(g) = T_G(f)/||f||∞. This keeps the ratio the same, so the supremum over all nonzero f equals the supremum over the unit sphere {f : ||f||∞ = 1}.

How does the inequality |∫ Gf| ≤ ∫ |G||f| lead to boundedness?

Start with |∫_0^1 G(x)f(x)dx|. Moving the absolute value inside gives |∫ Gf| ≤ ∫ |G(x)f(x)|dx = ∫ |G(x)||f(x)|dx. If ||f||∞ = 1, then |f(x)| ≤ 1 for every x, so ∫ |G(x)||f(x)|dx ≤ ∫ |G(x)|dx. Since |G| is continuous on [0,1], the integral ∫_0^1 |G(x)|dx is finite, so |T_G(f)| is uniformly bounded.

Why is the condition “G has no zeros” needed for the norm calculation?

The constructed test function H(x) = \overline{G(x)}/|G(x)| requires division by |G(x)|. If G(x0) = 0 for some x0, then |G(x0)| = 0 and H would be undefined (or not bounded) at that point. The no-zeros assumption guarantees H is well-defined and continuous on [0,1], with ||H||∞ = 1.

How does the complex conjugation in H ensure the integral becomes ∫ |G|?

With H(x) = \overline{G(x)}/|G(x)|, the product G(x)H(x) equals G(x)\overline{G(x)}/|G(x)| = |G(x)|^2/|G(x)| = |G(x)|. Taking absolute values then yields |T_G(H)| = |∫_0^1 |G(x)|dx| = ∫_0^1 |G(x)|dx because the integrand is nonnegative.

What shows the operator norm is not just an upper bound but the exact value?

The upper bound comes from |T_G(f)| ≤ ∫_0^1 |G(x)|dx for all ||f||∞ = 1. The lower bound comes from exhibiting one specific f (namely H) with ||H||∞ = 1 such that |T_G(H)| = ∫_0^1 |G(x)|dx. Since the supremum cannot be smaller than any achieved value and cannot exceed the universal upper bound, equality follows: ||T_G|| = ∫_0^1 |G(x)|dx.

Review Questions

Given T_G(f)=∫_0^1 G(x)f(x)dx on (C([0,1]),||·||∞), derive the inequality ||T_G|| ≤ ∫_0^1 |G(x)|dx.
Why does choosing H(x)=\overline{G(x)}/|G(x)| force equality in the operator norm?
What would break in the argument if G had a zero inside [0,1]?

Key Points

1
Define T_G on C([0,1]) by T_G(f)=∫_0^1 G(x)f(x)dx and measure inputs with the supremum norm.
2
Scaling reduces the operator norm computation to functions with ||f||∞=1.
3
The bound |T_G(f)| ≤ ∫_0^1 |G(x)|dx follows from |∫ Gf| ≤ ∫ |G||f| and |f(x)|≤1.
4
Continuity of G ensures ∫_0^1 |G(x)|dx is finite, so T_G is bounded.
5
The no-zeros condition on G makes H(x)=\overline{G(x)}/|G(x)| well-defined and guarantees ||H||∞=1.
6
Plugging H into T_G turns the integrand into |G(x)|, giving |T_G(H)|=∫_0^1 |G(x)|dx.
7
Combining the universal upper bound with the achieved lower bound yields ||T_G||=∫_0^1 |G(x)|dx.

Highlights

The operator norm of the integral functional f ↦ ∫_0^1 G(x)f(x)dx equals ∫_0^1 |G(x)|dx.

A sharp test function H(x)=\overline{G(x)}/|G(x)| saturates the supremum and forces equality.

Boundedness drops out immediately from |∫ Gf| ≤ ∫ |G||f| plus the constraint ||f||∞=1.

Topics

Operator Norm
Integral Functional
Supremum Norm
Bounded Operators
Test Function