If $T:X\to Y$ is a bijective bounded linear operator and $X$ is Banach then $T$ is invertible

functional-analysislinear-transformationsvector-spaces

Let $X$ and $Y$ be normed linear spaces and $T:X\to Y$ a bounded linear operator.

Notation:

  • $X'$ denote the dual space of $X$, and $T'$ the dual map of $T$.
  • If $W\subset X$, then ${}^\circ W=\{f\in X'\mid f(x)=0,\forall x\in W\}$ is the annhilator of $W$
  • If $Z\subset X'$, then $Z^\circ=\{x\in X\mid f(x)=0,\forall f\in Z\}$ is the annhilator of $Z$.

Then I need to prove that

  1. $\text{im}\,T$ is closed iff $\text{im}\,T={}^\circ\ker T'$
  2. $\ker T'=\{0\}$ iff $\text{im}\,T$ is dense in $Y$
  3. Let $X$ be Banach. Show that if $\inf_{x\neq 0}\frac{||Tx||}{||x||}>0$ and $\ker T'=\{0\}$ then $T$ is bijective and boundedly invertible.

I proved the first two parts, and I managed to show that $T$ is bijective, but I'm not sure how to prove that it has a bounded inverse.

Best Answer

If $X,Y$ are Banach then the inverse of a bijective bounded operator is automatically bounded, it is a consequence of the open mapping theorem. If $X$ or $Y$ is not Banach there are counter-examples so in your case, $a$ $priori$, you cannot apply this result.

You can show by hand that $T^{-1}$ is continuous, let $$ c = \inf_ {x \in X \setminus \{ 0 \}} \frac{||Tx||}{||x||} > 0, $$ as $T^{-1}$ is linear it is continuous if and only if there is a constant $c' > 0$ such that $$ \forall y \in Y,\quad ||T^{-1}y|| \leq c'||y||. $$ But forall $y \in Y$ not null there is $x \in X$ not null such that $y = Tx$ and $$ ||T^{-1}y|| = ||x|| \leq \frac{1}{c}||Tx|| = \frac{1}{c}||y||. $$

Actually there is another way to answer your question that is more "functionnal analysis", we can show that $Y$ is Banach. Indeed if $(y_n) = (Tx_n)$ is a Cauchy sequence of $Y$ by definition of $c$ $$ ||y_{n+p} - y_n|| = ||Tx_{n+p} - Tx_n|| = ||T(x_{n+p}-x_n)|| \geq c ||x_{n+p} - x_n|| $$ so $(x_n)$ is Cauchy and converges because $X$ is Banach.

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