Let $X$ be a Banach space and $T:X\rightarrow X'$, where $X'$ denotes the dual space of $X$. Let $T^*:X''\rightarrow X'$ be the adjoint of $T$. I am asked to show that if Ker $T\subset$ Ker $T^*$ (in the sense of the usual embedding from $X$ to $X''$) and range $T$ is closed, then there exists a constant $C>0$ such that
$\forall x\in X: \langle Tx,x\rangle_{X'\times X}\geq -C\|Tx\|_{X'}^2$
Any help would be appreciated. Thanks in advance!
Edit: fixed the type-o (range of $T^*$).
My attempt: Sadly, I don't have much of an idea on how to approach the problem (beyond really basic ideas like: if there exists $x_n$ such that $\|Tx_n\|=1$ and $\langle Tx_n,x_n\rangle < n$). I was able to prove that the inequality implies the inclusion between kernels by contradiction, showing $\langle Tz,x\rangle=0$ via parametrization $z=tz_0$ (for suitable $z_0$) and letting $t\rightarrow 0$. But this method is too indirect to try to somehow apply to address the other implication.
sketch of a proof:
Your idea with $\|Tx_n\|=1$ and $\langle Tx_n,x_n\rangle < -n$ was already a good idea.
Here is an idea how to continue based on the open mapping theorem. Let $Y$ denote the (closed) range of $T$. If we consider $T:X\to Y$, then $T$ is surjective and we can apply the open mapping theorem. Then we can find $y_n\in X$ such that $Ty_n=Tx_n$ but such that $y_n$ is bounded by a constant.
Since $T(y_n-x_n)=0$, we also have $T^*(y_n-x_n)=0$ or $T^*y_n=T^*x_n$. This can be used to show $\langle Tx_n,x_n\rangle =\langle Ty_n,y_n\rangle$. Since $y_n$ is bounded, this is a contradiction to the inequality $\langle Tx_n,x_n\rangle < -n$.