Let $X, Y$ be Banach spaces. If $T^\prime$ is an isomorphism so is $T$

Question

I spent quite a long time trying to proof this, but I am not 100% sure this proof works, and was hoping if someone could maybe proof read this for me.

Let $T^\prime: Y^\prime \rightarrow X^\prime$ the adjoint of $T$, show that if $T^\prime$ is an isomorphism so is $T$

My attempt: Because $T^\prime: Y^\prime \rightarrow X^\prime$ is an isomorphism $(T^\prime)^\prime: (X^\prime)^\prime \rightarrow (Y^\prime)^\prime$ is an isomorphism.
We consider the canonical embedding $J_X$ from $X$ into $(X^\prime)^\prime$ with $J_X(x) = ev_x$
$ev_x(\alpha) = \alpha(x)$
Because $X$ and $X^\prime$ are Banach it follows that $R(J_X)$ is closed. The same follows for $J_Y$

Consider $g = (T^\prime)^\prime$ from $R(J_X) \rightarrow (Y^\prime)^\prime$
Under this map with $ev_x \in (T^\prime)^\prime, \alpha \in Y^\prime$:
$g(ev_x)(\alpha) = (T^\prime)^\prime(ev_x)(\alpha) = ev_x(T^\prime(\alpha)) = T^\prime(\alpha)(x) = \alpha(T(x)) = ev_{T(x)}(\alpha)$ \

$\Rightarrow g(x) = ev_{T(x)}$
$\Rightarrow g(R(J_X)) \subset R(J_Y)$

Therefore the function $f = J_Y^{-1} \circ (T^\prime)^\prime \circ J_X$ is well defined and it holds:
$f(x) = T(x) \Rightarrow f = T$
$\Rightarrow T(X) = R(T) = J_Y^{-1} \circ g(R(J_X))$
And because $J_Y$ and $g$ are isomorphisms $R(T)$ is closed.

Because $T^\prime$ is injective $N(T^\prime) = \{0 \}$
$\Rightarrow$ by the closed range theorem $R(T) = (N(T^\prime))^\perp = Y$
$T$ can be written as a composition of injective maps $\Rightarrow T$ is injective.
$\Rightarrow T$ is bijective
From the inverse mapping Theorem it follows that $T^{-1}$ is continuous .
$\Rightarrow T$ is an isomorphism.

Answer 1

Your proof is correct, and it contains a few pieces that are important in their own right. For example, $T''\circ J_X = J_Y\circ T$. In fact, among my colleagues, we would typically not even distinguish between $X$ and $J_X(X)\subset X''$, so I would summarize this relationship as $T''|_X=T$.

There may be a simpler way to go about this.

A linear operator $S:E\to F$ is an isomorphic embedding (that is, an isomorphic embedding onto its range, which is then necessarily closed) if and only if there exist $0<a\leqslant A<\infty$ such that $$a\|e\|\leqslant \|Se\|\leqslant A\|e\|$$ for all $x\in X$. The inequality $a\|e\|\leqslant \|Se\|$ implies $\ker(S)=\{0\}$.

The injective associate of a bounded, linear operator $S:E\to F$ is the operator $\overline{S}:E/\ker(S)\to F$ given by $\overline{S}(e+\ker(S)=Se$. If $Q:E\to E/\ker(S)$ is the quotient map, then $S=\overline{S}Q$.

A bounded, linear operator $S:E\to F$ is closed range if and only if there exists $0<A<\infty$ such that for all $f\in S(E)$, $$\inf\{\|e\|:Qe=f\}\leqslant A\|f\|,$$ which is the same as saying $\overline{S}:E/\ker(S)\to F$ is an isomorphic embedding.

Last, $S:E\to F$ is an isomorphism if and only if it is an isomorphic embedding and dense range. Note that isomorphic embedding implies closed range, so dense range and isomorphic embedding together imply surjectivity.

If $T':Y'\to X'$ is an isomorphism, then so is $T'':X''\to Y''$, which means there exist $0<a\leqslant A<\infty$ such that $a\|x''\|\leqslant \|T''x''\|\leqslant A\|x''\|$ for all $x''\in X$. Using $T''\circ J_X=J_Y\circ T$ (or, more transparently but less formally, $T''|_X\equiv T$), we deduce that $$a\|x\|\leqslant Tx\|\leqslant A\|x\|$$ for all $x\in X$. Thus $T$ is an isomorphic embedding. We next use that $$\overline{T(X)}=\ker(T')_\perp=\{y\in Y:(\forall y'\in \ker(T'))(y'(y)=0)\}$$ together with $\ker(T')=\{0\}$ to get $\overline{T(X)}=\{0\}_\perp=Y$. So $T$ is dense range.

Just for the sake of completeness, let's see why $a\|x\|\leqslant \|Tx\|\leqslant A\|x\|$ implies that $T$ is injective and closed range. If $Tx=0$, then $\|x\|\leqslant \|Tx\|/a=0$, so $x=0$, and $T$ is injective. Assume $(Tx_n)_{n=1}^\infty$ is convergent to some $y\in Y$. Then $(Tx_n)_{n=1}^\infty$ is Cauchy, and $(x_n)_{n=1}^\infty$ is, too, since $$\|x_m-x_n\|\leqslant \|Tx_m-Tx_n\|/a,$$ which vanishes as $\min\{m,n\}\to \infty$. So $x=\lim_n x_n$ exists, and $y=Tx$ is in the range of $T$. Hence the range of $T$ is closed.