Injective operator on a Banach space

Question

Let $X$ be a Banach space, $T \in B(X)$ and $||I−T||=γ<1$. Show that $T$ is injective. My attempt:

Suppose that $Tx=Ty$, for some $x,y\in X$. Then, by the linearity of $T$, $$T(x-y)=0\implies ||T(x-y)||=0.$$ By definition of the operator norm, $$||T(x-y)|| \le ||T||\,||x-y||.$$ How can I show that this implies $||x-y||=0$?

Answer 1

The space of bounded linear operators on $X$, denoted $\mathcal{B}(X)$, is a Banach space, since $X$ is complete. The norm associated with $\mathcal{B}(X)$ is the operator norm, i.e. \begin{align*} \|T\| = \sup_{\|x\| = 1} \|Tx\|. \end{align*}

Recall that a normed vector space is complete if and only if every absolutely convergent series converges.

So consider the series \begin{align*} \sum_{n=0}^{\infty} \|I-T\|^n \end{align*} Since $\|I-T\| < 1$, this series converges, hence \begin{align*} \sum_{n=0}^{\infty} (I-T)^n \end{align*} converges in $\mathcal{B}(X)$ to some element, call it $S$. To show that $S = T^{-1}$, it suffices to show that $ST= I$. Observe, \begin{align*} ST &= \sum_{n=0}^{\infty} (I-T)^n T \\ &= \sum_{n=0}^{\infty} (I-T)^n - \sum_{n=0}^{\infty} (I-T)^{n+1} \\ &= \sum_{n=0}^{\infty} (I-T)^n - \sum_{n=1}^{\infty} (I-T)^{n} \\ &= (I-T)^0 \\ &= I \end{align*} Thus, $S = T^{-1}$. In particular, $T$ is injective.

Answer 2

It is enough to show that the kernel of $T$ is trivial. Suppose $T(x)=0$ for some vector $x\ne 0$. Then we have:

$1>||I-T||\geq \frac{||(I-T)(x)||}{||x||}=\frac{||I(x)-T(x)||}{||x||}=\frac{||x||}{||x||}=1$

Which is of course a contradiction.

Edit: Note that you don't really need completeness for this. On the other hand, user474986 proved a stronger statement in his answer, that $I-T$ is actually invertible in $B(X)$. For that you really need $X$ to be a Banach space.

Answer 3

Assume that $T$ is not injective, then there is some $x \in X$ with $\Vert x \Vert = 1$ s.t. $Tx = 0$ and thus

$$ \Vert I - T \Vert \geq \Vert (I - T)x \Vert = \Vert x - 0 \Vert = 1 $$

which is a contradiction.

Answer 4

You cannot show that $\| x - y\| = 0$ from the inequality $\|T(x-y)\| \le \|T\| \| x - y\|$.

However, you can show that the operator $I - T$ is a contraction, which means that there exists a unique fixed point of the operator $I-T$, which is $0$. Now suppose that $Tx = Ty$, then \begin{align*} (I-T)(x-y) = x - Tx - y + Ty = x - y + (Ty - Tx) = x -y. \end{align*} Thus also $x-y$ is a fixed point of $I-T$, which means that $x - y = 0$, or $x = y$ and hence $T$ is injective.