Continuity of Linear Maps

Real Analysis · Axiom Academy

EXAMPLE Continuity of Linear Maps A linear map is continuous exactly when it is bounded — we prove it, then read off . Let be a linear map between normed spaces. Show that T is continuous if and only if it is bounded — meaning there is a constant with for every x . Then make it concrete with the scaling map T(x) = 3x on , and find the that the definition of continuity demands. Pick how close the outputs must be (the green window of half-width ε). Shrink the input window by the stretch factor M = 3, i.e. set δ = ε/3, and every input within δ of a lands inside the green window. Static figure — no animation. Nice work. You proved a cornerstone of analysis: for a linear map, the algebra of a bound and the topology of continuity are the same fact. Bounded ⟺ continuous: for a linear T , for all x is equivalent to continuity (in fact to continuity at the single point 0 ). Linearity is the trick: T(x) - T(a) = T(x-a) turns "outputs close" into "one bound on a single vector," which is why works everywhere at once — T is uniformly continuous, even Lipschitz. Finite dimensions are automatic: on a finite-dimensional space every linear map is bounded, hence continuous — no exceptions. The concrete answer: for T(x) = 3x the bound is M = 3 , so (e.g. gives ). The "if and only if" matters: in infinite -dimensional spaces unbounded linear maps exist — and those are exactly the discontinuous ones. Boundedness is the whole story.

This is the written version of the interactive lesson above. See the full Real Analysis course.