# Prove that $Hom(V,W)\neq L(V,W)$

Let $V$ a normed space of infinite dimension and let $W\neq 0$ a normed space. Prove that $Hom(V,W)\neq L(V,W)$.

Let $\phi:V\to\mathbb{K}$ (where $\mathbb{K}$ is the base field) be a discontinuous linear functional$^{(1)}$. Let $x\in W\setminus 0$. Define $T:V\to W$ by $Tv=\phi(v)x$.

Let's show that $T$ is discontinuous: Since $\phi$ is discontinuous, it is unbounded, so there exists a bounded sequence of vectors $\{v_n\}$ in $V$ such that $\{\phi(v_n)\}$ is unbounded. But then, since $x\neq 0$, $\Vert x\Vert>0$, and it follows that $\{T(v_n)\}=\{\phi(v_n)x\}$ is unbounded in $W$. Thus $T$ is unbounded, i.e., discontinuous.

Thus we found $T$ which is linear and discontinuous, i.e., $T\in Hom(V,W)\setminus L(V,W)$.

$(1)$For example, let $B$ be a basis of $V$. Choose an infinite subset $\{x_n\}\subseteq B$, and define $\phi(x_n)=n\Vert x_n\Vert$ and $\phi(x)=0$ for $x\in B\setminus\{x_n\}$ (and extend it linearly to all of $V$). To see that $\phi$ is discontinuous, note that the sequence $x_n/n(\Vert x_n\Vert)$ converges to $0$ but $\phi(x_n)=1$.

The existence of discontinuous linear functions depends on the axiom of choice. See the question On every infinite-dimensional Banach space there exists a discontinuous linear functional.

• why did you edit $W\neq\emptyset$ for $W\neq0$? – Ariel Marcelo Pardo Jul 5 '15 at 13:09
• @ArielMarceloPardo It doesn't make sense to say that a vector space is empty (as a set), since one of the axioms states that there exists a zero vector. The correct condition is that $W$ is non-zero (i.e., it is contains a vector different from the zero vector). – user160185 Jul 5 '15 at 20:53