Proving Isomorphism of Linear Operator with ||A|| < 1

[Jaggis]

Hi, I have some trouble with the following problem:

Let E be a Banach space.

Let A ∈ L(E), the space of linear operators from E.

Show that the linear operator φ: L(E) → L(E) with φ(T) = T + AT is an isomorphism if ||A|| < 1.

So the idea here is to use the Neumann series but I can't really figure out how to apply it here. Any help?

 

[S.G. Janssens]

Let $A \in L(E)$, the space of linear operators from E.

This should be the space of bounded linear operators, I suppose?

So the idea here is to use the Neumann series but I can't really figure out how to apply it here. Any help?

It is part of the Neumann series theorem that if $X$ is Banach and $K \in L(X)$ satisfies $\|K\| < 1$, then $\phi := I + K$ has a bounded inverse. Using this, start by taking $X = L(E)$. How should $K$ be defined? Why does it then satisfy $\|K\| < 1$?

 

[Jaggis]

This should be the space of bounded linear operators, I suppose?

Oh, yes. You are right. My bad.

How should $K$ be defined? Why does it then satisfy $\|K\| < 1$?

Because I + K has a bounded inverse that can be represented as an infinite sum (according to the Neumann series theorem)

∑||K||^k,

we need ||K|| <1.

If φ(K) = T + TK, would the solution have something to do with the following reasoning: T + TK = T(I + K)? Now I +K has an inverse but T necessarily doesn't. So, how can I deduce that T(I +K) is also an isomorphism?

 

[S.G. Janssens]

You are looking at it the wrong way, it is irrelevant whether $T$ has an inverse or not. You are interested in invertibility of an operator (namely, $\phi$) on $X = L(E)$, not in invertibility of an operator on $E$ itself.

Note that if you choose $K$ correctly (you didn't specify it yet), then your $\phi$ and my $\phi$ are the same.

 

[Jaggis]

You are looking at it the wrong way, it is irrelevant whether $T$ has an inverse or not. You are interested in invertibility of an operator (namely, $\phi$) on $X = L(E)$, not in invertibility of an operator on $E$ itself.

OK. I see your point. I think.

Then I suppose that we want to use the Neumann series for φ. Because we demand that φ is an isomorphism, we need || φ-I || < 1.

As in

|| φ-I || = || T+KT - I || <1.

Now it remains to show that this holds if ||K|| <1?

Note that if you choose $K$ correctly (you didn't specify it yet), then your $\phi$ and my $\phi$ are the same.

But can we choose K freely? The way I understand it, the task is to show that

K ∈ L(L(E)) with ||K|| < 1 ⇒ φ: L(E) →L(E) with φ(T) = T + KT is an isomorphism

holds.

 

[S.G. Janssens]

From the OP I thought the task is to show that $\phi \in L(X)$ defined by
$$
\phi(T) := T + AT \qquad \forall\,T \in X := L(E)
$$
is an isomorphism whenever $A \in X$ satisfies $\|A\| < 1$. Isn't this what you wanted?

Assuming this is correct, what I was hinting at, is that given $A \in X$ you define $K \in L(X)$ by
$$
KT := AT \qquad \forall\,T \in X
$$
(So note: $K$ is not equal to $A$! The latter is in $X$, the former is in $L(X)$.) Then it follows that $\phi = I_X + K$ with $I_X$ the identity on $X$. Now you just have to prove that $\|K\| < 1$, which is trivial. (In fact, the whole exercise is rather easy, and maybe you are overthinking it? Just be sure to keep a clear distinction between $E$, $L(E)$ and $L(L(E))$. To help myself with this, I introduced the symbol $X$.)

 

[Jaggis]

OK, thanks for your help.