I am trying to prove the following statements: Let $X$ and $Y$ be normed spaces (not necessarily complete) Let $T\in L(X,Y)$ (meaning $T:X\to Y$ is a bounded linear map). Let $T^*:Y^*\to X^*$ denote the adjoint operator. Then:

  1. $T^*$ is sujective if and only if $T$ is an isomorphism;
  2. $T$ is surjective if and only of $T^*$ is an isomorphism.

Here an "isomorphism" $X\to Y$ is an injective linear operator $T:X\to Y$ such that there exists $c_1,c_2>0$ with $$c_1\|x\|\leq \|Tx\|\leq c_2\|x\| \text{ for all } x\in X.$$ In particular we do not require $T$ to be surjective.


I have proved statement 1. and I have also proved that if $T^*$ is an isomorphism then $T$ is surjective, and that if $T$ is surjective then $T^*$ is injective. An ideal next step would be to show that $Image(T^*)$ is closed in $X^*$, at which point I could apply the open mapping theorem to conclude that $T^*$ is an isomorphism (since I already know $T^*$ is a bounded operator). However, I am struggling to show that the image of $T^*$ is closed. Any ideas would be appreciated.

  • $\begingroup$ Make sure you are using the right definition of norm in the space $X^*$. In particular, $||T^* y^* || = \max_x \frac{||T^* y^* x||}{||x||} =\max_x \frac{|| y^* T(x)||}{||x||} \le || y^*|| \max_x \frac{||Tx||}{||x||} \le c_2 ||y^*||$. $\endgroup$ Nov 3, 2015 at 7:14
  • $\begingroup$ @daw yes and both spaces are complete since $X^*$ and $Y^*$ are always Banach spaces, and if $Image(T)$ is closed so is it. $\endgroup$
    – CWsl2
    Nov 3, 2015 at 7:24
  • $\begingroup$ @RomeoAlexander Hi Romeo, I can confirm I am using the right definition of the norm, and this computation seems to show that $T^*$ is bounded? It is the other direction that I am struggling with though $\endgroup$
    – CWsl2
    Nov 3, 2015 at 7:27
  • $\begingroup$ Hi Curtis. See my answer below. The most difficult part is guaranteeing the existence of a maximizer $z$. As daw mentions, the result it not true if we don't have completeness. $\endgroup$ Nov 3, 2015 at 8:04
  • $\begingroup$ This is a very strange definition of isomorphism. IMO you should call that property something else (bi-Lipschitz, for example...). $\endgroup$ Nov 3, 2015 at 8:21

3 Answers 3


The claim that $T$ is surjective implies range of $T^*$ being closed is not true.

Take $X=c_{00}=Y$ the space of sequences with finite length. The dual space can be identified with $l^1$. Define $$ Tx = (x_1, x_2/2, \dots, x_n/n,\dots). $$ Clearly, $T:X\to Y$ is injective and surjective, however $T^{-1}$ is not bounded.

Let $g\in l^1$ be given. Then for $x\in c_{00}$, $$ (T^*g)(x)=g(Tx) = \sum_k g_k x_k/k = \sum_k g_k x_k/k = h(x), $$ where $T^*g=h = (g_1,g_2/2,\dots)$. Now take $e_k=(0,\dots,0,1,0,\dots)$ with the non-zero entry at position $k$. Then $$ \|T^*e_k\|_{l^1} = \frac1k, \ \|e_k\|_{l^1}=1, $$ which shows that the claim is not true in the general situation. (The example is also valid for $Y=c_0$, which is a Banach space)

If in addition $X$ and $Y$ are complete, then the claim follows with the closed range theorem.


Let us answer the first question, hopefully the other one is similar enough:

If $T^{\ast}$ is surjective, then it is an open map, and so $\exists r>0$ such that $$ B(0,r) \subset T^{\ast}(B(0,1)) $$ where these denote the open balls in their respective spaces. Hence if $\varphi \in X^{\ast}, \|\varphi\| = 1, \exists \psi \in B(0,1)$ such that $$ \frac{r}{2}\varphi = T^{\ast}(\psi) $$ Then for any $x\in X$, $$ \frac{r}{2}|\varphi(x)| = |\psi(Tx)| \leq \|\psi\|\|T(x)\| < \|T(x)\| $$ Varying over all $\varphi$ as above, we see that $$ \frac{r}{2}\|x\| \leq \|T(x)\| $$ and so $T$ is an isomorphism.

Conversely, if $T$ is an isomorphism, then if $\varphi \in X^{\ast}$, then define $\psi : \text{Im}(T) \to \mathbb{C}$ by $$ \psi(Tx) := \varphi(x) $$ and this is well-defined since $T$ is injective. Furthermore $$ |\psi(Tx)| \leq \|\varphi\|\|x\| \leq \frac{\|\varphi\|}{c_1}\|T(x)\| $$ and so $\psi$ is bounded on $\text{Im}(T)$, and so it extends to a bounded linear functional $\psi \in Y^{\ast}$. Now clearly, $$ T^{\ast}(\psi) = \varphi $$ and hence $T^{\ast}$ is surjective.

  • 2
    $\begingroup$ The other question is not similar: While $T^*$ is naturally a mapping between Banach spaces, this does not hold for $T$.... $\endgroup$
    – daw
    Nov 3, 2015 at 9:16

Fix $y \in Y$. Let $z$ be a unique point in $Y$ such that $||y^*|| = \frac{||y^* z||}{||z||}$, ie. $$ \max_w\frac{||y^*w||}{||w||} = \frac{||y^* z||}{||z||}. $$ (Note that we need completeness of $Y$ to be assured of the inclusion of $z$ in $Y$.) Since $T$ is surjective, there exists a $\tilde x \in X$ such that $T \tilde x = z$. Therefore $$ ||T^* y^*|| = \max_x \frac{||y^* T x||}{||x||} \ge \frac{||y^* T \tilde x||}{||\tilde x||} = c_1\frac{||y^*T \tilde x||}{||T\tilde x||} = c_1 \frac{|| y^*z||}{||z||} = c_1 || y^*||. $$ where we have used the inverse of the inequality $c_1 ||x|| \le ||T x||$ for general $x$. The existence of such a constant $c_1> 0$ requires the completeness of $X$.

On the other side, we have $$ ||T^* y^* || = \max_x \frac{||T^* y^* x||}{||x||} =\max_x \frac{|| y^* T(x)||}{||x||} \le || y^*|| \max_x \frac{||Tx||}{||x||} \le c_2 ||y^*||, $$ where we have used the Cauchy-Schwarz inequality.

So $c_1 ||y^*|| \le ||T^* y^*|| \le c_2 ||y^*||$.


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