I am reading Kreyszig's Introduction to Functional Analysis.
He explains in an example why the differentiation operator is not a bounded operator. I cannot follow his explanation. I sometimes have difficulties wrapping my head around taking norms in function spaces.
He describes.
Let $x_n(t) = t^n$, where $n \in \mathbb{N}$. Then $\Vert x_n \Vert = 1$ and $Tx_n(t) = x_n' (t) = nt^{n-1}$, so that $\Vert Tx_n \Vert = n$ and $\frac{\Vert Tx_n\Vert}{x_n}=n$.
The norm should be $\Vert T \Vert = \underset{\substack{x \in D(T) \\ x \neq 0}}{\sup} \frac{\Vert Tx\Vert}{\Vert x\Vert}$
Why is $\Vert x_n \Vert = 1$? Is there any good advice on how to think about taking the norm in function spaces to not get confused?
Happy new year!
Any polynomial can understood as a vector in a vector space. In this case the author implicitly chooses the monomial basis. So with the basis $(x^n)_n$ the vector $x^n$ is a vector which $n+1$th component is 1 and every other component is zero. This vector trivially has norm 1. Since something like $(0, 0, .... , 0, 1, 0, 0, ...)$ represents your monomial over the monomial basis and is in fact a unit vector.
The derivative of that polynomial is $nx^{n-1}$ which is $n$ in the $n$th component and zero everywhere else. So if you were trying to pick a bound for this derivative operator working on the basis of monomial i could always find a polynomial of sufficiently large degree to violate your bound.
If you were to choose a different basis such as $(\frac{x^n}{n!})_n$ this would no longer hold. However the linear operator which changes basis from $(\frac{x^n}{n!})_n$ to $(x^n)_n$ would be unbounded instead.