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Bounded Linear Operators and the Operator Norm

For a linear map between normed spaces, continuity and boundedness turn out to be the same condition. The operator norm records the worst stretching the map can perform on a unit vector.

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What a learner can do afterwards

  • Prove that a linear map is continuous exactly when it is bounded
  • Compute the operator norm of a multiplication or integral operator
  • Give an unbounded linear map and explain why differentiation is the standard example
  • State submultiplicativity for bounded operators: ||ST|| <= ||S|| ||T||.

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The operator norm of a linear map T is the least upper bound of ||Tx|| over all unit vectors x. T is bounded when this number is finite, and then ||Tx|| is at most ||T|| times ||x||. Bounded implies continuous, since ||Tx minus Ty|| is at most ||T|| times ||x minus y||. Conversely, if T is continuous at 0, some ball of radius r maps into the unit ball, and scaling gives ||T|| at most 1/r.

In a finite-dimensional space, fix a basis and T becomes a matrix whose columns are the images of the basis vectors. Then ||Tx|| is at most the sum of the column lengths times the largest absolute coordinate of x, and norm equivalence bounds that magnitude by a constant times ||x||. So every linear map there is bounded. Differentiation is different: on polynomials with the sup norm on [0, 1], x to the n has norm 1 but its derivative has norm n.

x1derivativeof x1x^31derivativex^33x^91derivativex^99
Each x to the n has sup norm 1, but its derivative has sup norm n, so no single bound works.
Try it together

Multiplication by g sends f to gf. On L2 of [0, 1], take g = 3x. Since |gf| <= 3|f|, the operator norm is at most 3. Functions concentrated near x = 1 are stretched by almost 3, so it is exactly 3. Rotations preserve length. Nonzero orthogonal projections never stretch and fix nonzero vectors in their range, so both have norm 1. Cauchy-Schwarz bounds a dot product by the product of lengths and helps estimate operator norms.

Chaining bounded maps multiplies their bounds at worst. For a unit vector x, ||STx|| is at most ||S|| times ||Tx||, which is at most ||S|| times ||T||. So ||ST|| is at most ||S|| ||T||, which is submultiplicativity.

Take the supremum over unit inputs. For linear maps, boundedness and continuity are equivalent. Test differentiation using x to the n.

Original lesson, questions and flashcards authored for this dataset.

Original notes and questions authored for this dataset. Enrichment flashcards were drafted from collected fact candidates. Includes video references by Khan Academy.

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Bounded Linear Operators and the Operator Norm · Mathematics, ages 23 to 24 · LightMySky