Continuity and Boundedness of Linear Maps

• categories: Functional analysis, Theorem, Linear algebra

For a Linear map $T:X\to Y$ between normed spaces $X$ and $Y$, the concepts of Continuity and boundedness are equivalent.

Theorem

A linear map $T:X\to Y$ is continuous if and only if it is bounded.

Intuition

For linear maps, continuity at any single point (usually the origin) implies continuity at every point due to linearity. Boundedness means there exists a constant $C\ge 0$ such that $\Vert T(x)\Vert \le C\Vert x\Vert$ for all $x\in X$. Thus, a bounded linear map cannot “blow up” and maps bounded sets in $X$ to bounded sets in $Y$.

Proof Outline

1. Boundedness implies continuity:

   • Suppose $T$ is bounded; then there exists $C\ge 0$ such that $\Vert T(x)\Vert \le C\Vert x\Vert$ for all $x\in X$.
   • For any $ϵ>0$, choose $\delta =ϵ/C$. Then, for $\Vert x\Vert <\delta$, we have: $\Vert T(x)\Vert \le C\Vert x\Vert <C\cdot \frac{ϵ}{C}=ϵ,$ showing that $T$ is continuous at $0$, and hence everywhere.

2. Continuity implies boundedness:

   • Assume $T$ is continuous at $0$. Then, there exists $\delta >0$ and $M>0$ such that $\Vert x\Vert <\delta$ implies $\Vert T(x)\Vert <M$.
   • For any $x\ne 0$, consider $y=\delta x/\Vert x\Vert$. Since $\Vert y\Vert =\delta$, we have $\Vert T(y)\Vert \le M$, leading to: $\Vert T(x)\Vert =\frac{\Vert x\Vert }{\delta }\Vert T(y)\Vert \le \frac{M}{\delta }\Vert x\Vert .$
   • Setting $C=M/\delta$, we find $\Vert T(x)\Vert \le C\Vert x\Vert$, showing that $T$ is bounded.