# If a closed linear operator is one to one on a dense set, then is it one to one on the whole set?

I am looking at the proof of the following proposition.

Here $A$ is a closed linear operator on $L$, that is, the graph of $A$ is closed subspace of $L \times L$. And $\lambda$ is said to belong to the resolvent set $\rho(A)$ of $A$, if $\lambda - A(\equiv \lambda I-A)$ is one-to-one, Range($\lambda - A)=L$, and $(\lambda - A)^{-1}$ is a bounded linear operator on $L$.

In the proof, as shown below, (2.6) shows that $\lambda-A$ is one-to-one on $\mathcal{D}(A)$, which is a dense subset of the vector space $L$ by a theorem preceding this. However, I don't see how this fact directly means that $\lambda - A$ is one-to-one on $L$. If a closed linear operator is one to one on a dense set, then is it one to one on the whole set? I would greatly appreciate any explanation on this line of the proof.

• Why do you need it to be injective on whole set? As $A$ is generator $\lambda - A$ is closed, and therefore for it to be invertible it only needs to be a bijection from $D(A)$ onto whole set. – user160738 Feb 21 '17 at 20:53
• @user160738 Isn't it required to define the inverse? – takecare Feb 21 '17 at 20:56
• No, only requirement is for it to be injective on $D(A)$. $U_{\lambda}$ takes care of both surjectivity and injectivity, and closedness of $\lambda -A$ implies it is invertible with $U_{\lambda}$ being the resolvent – user160738 Feb 21 '17 at 20:57
• @user160738 Can you explain a little more? I am confused. From the definition of the book, which I just posted above, we need $\lambda - A$ to be injective on the domain, which is $L$, but (2.6) only shows it for $D(A)$. Also, why does the closedness of $\lambda-A$ imply it is invertible with $U_\lambda$ being the resolvent? – takecare Feb 21 '17 at 21:01
• Domain of $\lambda - A$ is $D(A)$ for all $\lambda$, how can it be $L$? – user160738 Feb 21 '17 at 21:08

You have

$$D(\lambda - A)=\{x\in L:\lambda x - Ax \in L\} = \{x\in L : Ax\in L\}=D(A)$$

because $\lambda x - Ax$ lies in $L$ iff $Ax$ lies in $L$.

So you only need it to be injective on $D(A)$. $U_{\lambda}$ takes care of both its surjectivity and injectivity.

Finally, once that's done it is autumatically invertible. This is applies to slightly more general case:

Lemma: Suppose that $A$ is a closed linear operator on a Banach Space $X$, then it is invertible iff it is bijective onto $X$.

Clearly if it is invertible then it is bijective onto $X$. So the converse is interesting one;

Let $G(A)$ denote its graph, and suppose that $A$ is bijective onto $X$. Then its algebraic inverse exists, call it $B$. Then the map $(x,y)\mapsto (y,x)$ from $G(A)$ bijective onto $G(B)$ is a homeomorhpism, so $G(A)$ closed implies $G(B)$ is closed. Now as $X$ is complete, and $B$ is a bijection from $X$ onto $D(A)$, by closed graph theorem $B$ is bounded. So $A$ is invertible