Let $H$ be a Hilbert space, $B\colon H\times H\to C$ a sesquilinear form. Assume that there's $0 \leq M <\infty$ such that for all $x,y\in H$: $|B(x,y)| \leq M\|x\| \|y\|$.
A. Show that there is a single $T\in B(H)$ such that for every $x,y\in H$: $B(x,y)=(x,Ty)$ and $T$ satisfies $\|T\|\leq M$.
B. Starting from this part, assume that there's $0< m<\infty$ such that for every $x,y\in H$: $B(x,x)\geq m\|x\|^2$. Show that $T$ is injective.
C. Show that $\mathrm{Im}(T)$ is dense in $H$.
D. Show that $\mathrm{Im}(T)$ is closed in the following way: $\forall x\in H$ by the inequality $m\|x\|^2\leq B(x,x)=(x,Tx)$ deduce that $m\|x\|\leq \|Tx\|$, and use a suitable theorem regarding these operators. Another approach, see what you'll get if $(Tx_n)$ is a Cauchy series in $H$.
E. Deduce that $T$ is invertible and $\|T^{-1}\| \leq m^{-1}$.
In A, I know that we need to rewrite the given form, but did not het the result mentioned.
B: using the information given in this part we have, $(x,Tx)=B(x,x)\geq m\|x\|^2$, now assume that $Tx=0$ so we have $0=(x,0)\geq m\|x\|^2\geq 0$ (since $m>0$), So $x$ must be $0$.
C: by contradiction, assume that $\mathrm{Im}(T)=TH \subset H$ and there is $u\in H\setminus TX$, then by the Hahn banach theorem/claim there exists a linear bounded functional $x^*\neq 0 \in H^*$ such that $x^*(TH)=0$ so $T^*x^*H=0$ ,therefore $T^*x^*=0$ then $x^*=0$ in contradiction.
D. $m\|x\|^2\leq B(x,x)=(x,Tx)\leq \|x\| \|Tx\|$, so $m\|x\|\leq \|Tx\|$. Now, using a theorem about an operator $T$ that satisfies this inequality, we get that $T$ has a close range. How to do this part in the other suggested method?
E: From parts B we have that $T$ is injective and from C and D, we got that $\mathrm{Im}(T)$ is closed and dense in $H$ and therefore $T$ is surjective. In A we have that $T\in B(H)$, so by the inverse bounded theorem we deduce that $T$ has a bounded inverse.
Can you please tell me how to do part A, and the remaining part in E, and if what I did is okay.
Thanks
Hints:
For part A, use the Riesz representation theorem. For any given $\ y\in H\ $ the function $\ \phi_y:H\rightarrow\mathbb{C}\ $ defined by $$ \phi_y(x)=B(x,y) $$ is a bounded linear functional on $\ H\ $ $\big($since $\ \big|\phi_y(x)\big|\le M\|y\|\|x\| $ for all $\ x\in H\ \big)$. Therefore, by the Riesz representation theorem, there exists a unique $\ z_y\in H\ $ such that \begin{align} \phi_y(x)&=\big(x,z_y\big)\\ &=B(x,y) \end{align} for all $\ x\in H\ $. Define $\ T:H\rightarrow H\ $ by putting $\ Ty=z_y\ $. Use the identity and inequality $$ (x,Ty)=B(x,y)\le M\|x\|\|y\|\ , $$ and the sesquilinearity of $\ B\ $ and $\ (.,.)\ $ ,to establish that $\ T\in B(H)\ $ with $\ \|T\|\le M\ $.
For the remaining part of E $\big(\ \big\|T^{-1}\big\|\le m^{-1}\ \big)$, put $\ x=T^{-1}y\ $ in the inequality $\ m\|x\|\le \|Tx\|\ $ you've already established in part D.