I am working on this functional analysis exercise:
Is the following proposition true?
$X$ is a Banach space, and $A: X \rightarrow X^{*}$ be a bounded linear operator. Let $B=\left.A^{*}\right|_{X} .$ If $B-A$ is surjective, then there exists $M>0$ such that $$ \|x\|_{X} \leq M\|A x-B x\|_{X^{*}}, \text{for all }x \in X $$
I first want to determine the operator $C=B-A:X\to X^*$:
For every $f_v \in X^{**}$, $g\in X$, we have $\langle A^*v,g\rangle=\langle A^*f_v,g \rangle=\langle f_v,Ag\rangle=(Ag)v$. Therefore $(Bx)v=(Av)x$. Hence $C: X\to X^*,\ x\mapsto C_x\in X^*$ such that $C_xv=(Av)x-(Ax)v$.
I also know that the statement is true iff $C$ has an continuous inverse.
So how to prove if $C$ is invertible or not?
For $T : X\to X^*$ let us denote $T^+ = T^*|_X$. As you saw already, $(T^+x)y = (Ty)x$ for $x,y\in X$.
Now, if $T$ is surjective, then $T^+$ is injective. Indeed, let $T^+x = 0$. Then $(Ty)x = 0$ for all $y\in X$ and so $x^*(x) = 0$ for all $x^*\in X^*$, which implies $x=0$, since $$ \|x\| = \sup\{|x^*(x)| : x^*\in X^*,\,\|x^*\|=1\}. $$ For $x,y\in X$ we also have \begin{align} [(A^+-A)^+x]y &= [(A^+-A)y]x\\ &= (A^+y)x - (Ay)x\\ &= (Ax)y-(A^+x)y\\ &= [(A-A^+)x]y. \end{align} Thus $A^+-A = -(A^+-A)^+$ is injective and the claim is true by the bounded inverse theorem.