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We say that a bounded operator $\varphi:X\to Y$ is

  • $c$-topologically injective if $\Vert\varphi(x)\Vert\geq c\Vert x\Vert$ for all $x\in X$
  • $c$-topologically surjective if for all $y\in Y$ there exist $x\in X$ such that $\varphi(x)=y$ and $\Vert x\Vert\leq c\Vert y\Vert$

I have already proved that $$ \varphi \;\text{$c$-topologically injective}\Longleftrightarrow\varphi^*\; \text{$c^{-1}$-topologically surjective}\\ \varphi \;\text{$c$-topologically surjective}\Longrightarrow\varphi^*\; \text{$c^{-1}$-topologically injective} $$ and got stuck at proving that $$ \varphi \;\text{$c$-topologically surjective}\Longleftarrow\varphi^*\; \text{$c^{-1}$-topologically injective}\tag{1} $$

Edit:

Thank to rayuela, now I can perform the proof. Let $\varphi^*$ is $c^{-1}$ topologically injective. Assume that $\mathrm{cl}_Y(\varphi(cB_X))\not{\supset} B_Y$, then we have some $y_0\in B_Y\setminus\mathrm{cl}_Y(\varphi(cB_X))$. By geometric form of Hahn Banch theorem there exist $g\in Y^*$ and $\gamma_2>\gamma_1>0$ such that $$ \mathrm{Re}(g_0(y_0))>\gamma_2>\gamma_1>\mathrm{Re}(g_0(y)) $$ for all $y\in \mathrm{cl}_Y(\varphi(cB_X))$. By considering $g=2(\gamma_1+\gamma_2)^{-1}g_0$ we get $$ \mathrm{Re}(g(y_0))>1>\mathrm{Re}(g(y)) $$ for all $y\in \varphi(B_X)$. For all $\lambda\in\mathbb{C}$ with $|\lambda|<1$ and $y\in\varphi(B_X)$ we have $\lambda y\in\varphi(B_X)$, so $$ |g(y_0)|>\mathrm{Re}(g(y_0))>1>|g(y)| $$ for all $y\in \varphi(cB_X)$. Since $|g(y_0)|>1$ while $y_0\in B_Y$, then $\Vert g\Vert> 1$. On the other hand for all $x\in c B_X$ we have $|g(\varphi(x))|<1$, i.e. $|\varphi^*(g)(x)|<1$. So $\Vert\varphi^*(g)\Vert<c$ and from $c^{-1}$-topological injectivity of $\varphi^*$ we get $\Vert g\Vert\leq c\Vert\varphi^*(g)\Vert\leq 1$. Contradiction, so $B_Y\subset\mathrm{cl}_Y(\varphi(cB_X))$. By open mapping theorem $\varphi(cB_X)\supset B_Y$ which is equivalent to $c$-topological surjectivity of $\varphi$.

Thank you for taking time.

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See Rudin, Functional analysis, Lemma 4.13.

You were not so far off track: Instead of arguing by contradiction, argue directly. Let $A$ be the closure of $\varphi(B_X)$ and suppose $y_0 \in Y \setminus A$. Since $A$ is closed, convex and balanced we can apply Hahn-Banach and find a functional $g \in Y^\ast$ such that $$ \lvert \langle g, a \rangle\rvert \leq 1 \lt \lvert \langle g,y_0\rangle\rvert $$ for all $a \in A$. For $x \in B_X$ we have $\varphi(x) \in A$, so $\lvert \langle \varphi^\ast g,x \rangle \rvert = \lvert \langle g,\varphi(x)\rangle\rvert \leq 1$. Taking the supremum over $x \in B_X$ we deduce $1 \geq \lVert \varphi^\ast g\rVert \geq c^{-1} \lVert g \rVert$ where the last inequality follows from $c^{-1}$-topological injectivity. This is as far as you got.

But now $$ c\lVert y_0\rVert \geq \lVert g\rVert \lVert y_0\rVert \geq \lvert \langle g,y_0\rangle\rvert > 1 $$ and $\lVert y_0 \rVert \gt c^{-1}$. This means that $c^{-1}B_Y \subseteq A$ and now use completeness of $X$ and the usual Banach-Schauder back-and-forth argument in the proof of the open mapping theorem shows that $c\varphi(B_X) \supseteq B_Y$ which is what you want.


To see that you need completeness of $X$ consider the following: suppose $\varphi \colon X \to Y$ is $c$-topologically surjective. Restricting $\varphi$ to a dense subspace $D$ of $X$ will not change the adjoint: $D^\ast = X^\ast$ and $(\varphi|_D)^\ast = \varphi^\ast$. However, restricting will affect $c$-topologial surjectivity.

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  • $\begingroup$ Thank you! I relized mistake in my proof. So a small fix will salvage it. Also I'm gratefull to you for this simple counterexample. $\endgroup$ – Norbert Apr 5 '13 at 7:54

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