Banach Spaces, Bounded Linear Operator

Bounded Linear Operator

A linear operator is a map between vector spaces that respects addition and scaling. This definition comes right out of linear algebra. But what is a bounded linear operator?

Assume the domain and range are normed vector spaces. The operator f is bounded if there is some constant k such that |f(x)| ≤ k×|x|. In other words, the function does not grow faster than linear.

Note that f(0) has to be 0, but this is the case for any linear operator.

The Norm of an Operator

If f is bounded it has a norm, denoted |f|, which is the lower bound of all the constants k that make f a bounded operator. Can we home in on |f|?

If x satisfies our constraint for a fixed k, then so does cx. To see if k is valid, there is no need to test the multiples of x. It is enough to test x/|x|, the unit vector in the direction of x. Consider all the points x on the unit sphere and evaluate |f(x)|. (I'm calling it a sphere, but I really mean all the points that are a distance 1 from the origin. This could be the surface of a cube, or almost any other shape, depending on the norm.) Let k be the least upper bound, and f is a k bounded linear operator. Lower values of k will not do, thus |f| = k.

In Rn, when f is implemented as a matrix, you might think |f| is the largest eigen value, but this need not be the case. Let f be the 2×2 upper triangular matrix [1,1|0,1]. The eigen values are 1, but run 1,0 through the matrix and get 1,1 with length sqrt(2).

If f is a normal matrix, e.g. a symmetric matrix, then its eigen vectors are orthogonal, and |f| is indeed the largest eigen value. Of course we have to take the norms of the eigen values, so that -4 is bigger than 3, forcing k = 4. And in the world of complex numbers, 4+3i is larger still, driving k up to 5.

If f is S mod a closed subspace U, as described in the previous section, the bound on f is 1. Distances compress to 0 along some coordinates, and hold steady along others.

Compact Sphere

Assume f is continuous, and the unit sphere is compact. The norm is continuous, thanks to the triangular inequality, so |f(x)| is a continuous function on the unit sphere. This is a continuous function from a compact set into the reals. The image is compact, hence closed and bounded. Thus our linear operator is bounded.

Let's apply the above when the domain is Rn. The unit sphere is closed and bounded in Rn, hence compact. We only need show continuity. Focus on one of the n coordinates. Our linear operator is continuous on R; in fact it scales R by a fixed amount and embeds it in the range. A linear operator on Rn is the sum of n linear operators on R, and is continuous. So we have a continuous function on a compact set, and every linear operator on Rn is bounded.

Continuous iff Bounded

Let f be a linear operator. If f is bounded, say |f| = k, then f is uniformly continuous. Given two points x and y, let d = x-y. Now |f(d)| is bounded by k×|d|. By linearity, f(d) = f(x)-f(y). Thus distance is magnified by at most k, everywhere, and f is uniformly continuous.

Clearly a function that is continuous everywhere is continuous at 0. Let's complete the circle by showing continuity at 0 implies a bounded function. (Since f is linear, we could use any base point; 0 is merely convenient.)

Select an r so that |x| < r implies |f(x)| < 1. The norm of the image of the sphere of radius r is at most 1, hence 1/r acts as a bound for f.

Given a linear operator f on a normed vector space, f is continuous at a point, iff f is bounded, iff f is uniformly continuous everywhere.

Unbounded

It's easy to build a linear function that is not bounded, and not continuous. Let b1 b2 b3 etc form a basis for an infinite dimensional vector space. Let f map b1 to b1, b2 to 2b2, b3 to 3b3, b4 to 4b4, and so on. The image of the jth unit vector has length j, and f is unbounded.