# Existence of nontrivial bounded linear operator?

Are there two normed spaces such that there is no nontrivial bounded linear operator between them: $$\nexists T:X\to Y: T\text{ nontrivial, linear and bounded}$$

• Take a one-dimensional subspace of $Y$ and use a non-trivial bounded linear functional on $X$ to construct a non-trivial bounded operator from $X$ to this subspace. – David Mitra May 31 '14 at 15:10
• Only for $X = \{0\}$ or $Y = \{0\}$. – Daniel Fischer May 31 '14 at 15:11
• So there always exist nontrivial operators? – C-Star-W-Star May 31 '14 at 15:14
• Can you explain the construction? – C-Star-W-Star May 31 '14 at 15:15
• If you allow your norm to take value $\infty$, you could endow $Y$ with the (basically useless) norm $$\| y\| = \begin{cases} 0 & y=0_Y \\ \infty & y \neq 0_Y. \end{cases}$$ In this case, the only bounded linear operator is trivial as long as there is some $0_X \neq x \in X$ such that $\|x\|< \infty$. – Tom May 31 '14 at 15:18

The Hahn-Banach Theorem gives us a wealth of nontrivial bounded linear functionals on (nontrivial) normed linear space $X$. We can for example take the "coordinate" map on a one-dimensional subspace $\{ru | r \in \mathbb{R} \}$, $f(ru)=r$, and extend it to a map $F:X \to \mathbb{R}$ also of the same norm as $f$.

Then another nontrivial bounded map $G:\mathbb{R} \to Y$ can be defined by $G(r) = ry$ for any nonzero vector $y \in Y$. The composition $T = G \circ F$ is then a nontrivial bounded linear transformation from $X$ to $Y$.

• As answered in the Comments by David Mitra and Daniel Fischer. – hardmath May 31 '14 at 15:40
• This doesn't show that any normed spaces admit nontrivial bounded linear operators! – C-Star-W-Star May 31 '14 at 15:49
• Why doesn't it? – hardmath May 31 '14 at 15:53
• Oh I'm sorry I overread the last sentence... – C-Star-W-Star May 31 '14 at 16:31