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

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.

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 $$T''|_X=T$$.

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 $$0 such that $$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 $$0 such that for all $$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 $$0 such that $$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.

• Thank you very much! Nov 21 at 7:23