Identity Mapping on Normed Vector Space is Bounded Linear Operator

Theorem

Let $\Bbb F$ be a subfield of $\C$.

Let $\struct {X, \norm \cdot}$ be a normed vector space.

Let $I : X \to X$ be the identity mapping on $X$.

Then $I$ is a bounded linear operator.

Further:

Let $\map B X$, be the space of bounded linear operators on $X$.

Then:

$\norm I_{\map B X} = 1$

where $\norm {\, \cdot \,}_{\map B X}$ denotes the norm on $\map B X$.

Proof

Let $\lambda \in \Bbb F$ and $x, y \in X$.

Then, we have:

$\map I {\lambda x + y} = \lambda x + y = \lambda I x + I y$

So $I$ is a linear operator.

Further, we have:

$\norm {I x} = \norm x$

for each $x \in X$, so $I$ is bounded.

Further, for all $x \in X$ with $\norm x = 1$ we have that $\norm {I x} = 1$, so that:

$\set {\norm {I x} : \norm x = 1} = \set 1$

so that:

$\norm I_{\map B X} = 1$

$\blacksquare$