I have the following example in my lecture notes:
Let $X = C[0,1]$ be a vector space of continuous real functions on the interval $[0,1] $ with the sup norm and $Y=\mathbb{R}$ and let $T:X\to Y$ be an operator defined as $Tf=f(0)
\space\space\forall f \in X$
Prove that $T$ is a a bounded linear operator and find the norm if possible.
The lineary is inmediate. For the boundedness
$\|Tf\|_Y =|Tf|=|f(0)|\leq \max_{[0,1]}|f(t)| = \|f\|_X$
from which it follows that T is bounded and $\|T\|\leq 1$
In particular $\|T\|=1$, since for $f \equiv1$, $\|Tf\|_Y=|f(0)|=|1|=1=\|f\|_X$
My definition of norm of a bounded linear operator is :
Let $T:X \to Y$ be a bounded linear operator $\|T\|=\inf\{ C>0; \|Tx\|_Y \leq C\|x\|_X \space \forall x \in X \}$ is the norm of $T$
Can someone please help me understand why $\|T\|=1$ ? I fail to see how when the inequality in the definition becomes an equality for a particular value of $f$, the $\inf$ over the values of $C$ that defines the norm of the operator becomes equal to the value of the constant $C$.
Thanks in advance.
Best Answer
Suppose $\|Tf\| \leq C\|f\|$ for all $f$. Taking $f=1$ we get $1 \leq C$. Thus every element of the set $\{C: \|Tf\| \leq C\|f\|\, \forall f \}$ is greater than or equal to $1$. This implies that the infimum of the set is also greater than or equal to $1$.