Erwin Kreyszig's Introductory Functional Analysis With Applications, Problem 8, Section 2.7

Question

Here is Problem 8 in the Problem Set following Section 2.7 in Erwine Kryszeg's Introductory Functional Analysis With Applications:

Show that the inverse $T^{-1} \colon R(T) \to X$ of a bounded linear operator $T \colon X \to Y$, where $X$ and $Y$ are two normed spaces over the same field ($\mathbb{R}$ or $\mathbb{C}$) and $R(T)$ denotes the range of $T$, need not be bounded.

As a hint, Kryszeg suggest the following operator:

Let $T \colon \ell^\infty \to \ell^\infty$ be defined by $$ Tx \colon = \left(\frac{\xi_j}{j}\right)_{j=1}^\infty \, \, \, \forall x \colon= (\xi_j)_{j=1}^\infty \in \ell^\infty. $$

How to characterise the range of this operator?

And how to show that the inverse of this operator is not bounded?

Answer 1

The range of this operator is a subspace of $C_{0}$, which consisting of elements eventually go to zero. It can be characterized by $$ \{a_{i}\}\in l^{\infty}, \exists N\in \mathbb{N}, |a_{i}*i|\le C, \forall i\ge N $$ It is not difficult to see that the inverse map $$ \{a_{i}\}\rightarrow \{ia_{i}\} $$ is not bounded on the sequence $a_{i}=1,\forall i$ under $l^{\infty}$-norm.

Answer 2

How do you invert the map $(a_1,a_2,\cdots) \rightarrow (a_1,\,a_1/2,\,a_3/3,\,\cdots)$? Let

$b_n=(a_1,\,a_2/2,\,a_3/3,\cdots,\,a_k/k,\,\cdots)$ , how do you get the original a_n back?

Now, to see if $T^{-1}:= (a_1,\,2a_2,\,\cdots)$ is bounded, you need to know what the

norm is that is used in $l^{\infty}$ : It is the sup norm $\|a_n\|:=\sup_n |a_n| $

To check first that T is bounded , you need to show that there exists a finite Real value M with : $$\sup_{\|x\|=1} T(x)<M= \sup_{\|a_n\|=1} T(a_n)<M$$

For this, you first need to find out what the unit ball is in $Sup$ norm: it is the

collection of sequences with subsequences whose sup is $1$ , e.g., $ a_n = 1/2, 1/2+1/4,\,\cdots,\, (1/2+1/4+...+1/2^n),\,\cdots$

Now, to test the boundedness of $T^{-1}$ look for a sequence $A_n:= (a_1,\,a_2,\,\cdots,a_n,\,\cdots)$

and

send it to $T^{-1} (A_n)=(a_1,\,2a_2,\,3a_3,\,\cdots)$ Is it the case that the supremum of

$A_n$ always exists ( as a Real number )? Notice how the terms in the sequence

are multiplied by increasingly-large numbers. Find a sequence that converges but

does not decrease too fast.