In this question $F$ stands for a complex Hilbert space with inner product $\langle\cdot\;,\;\cdot\rangle$ and the norm $\|\cdot\|$. Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on $F$.
Let $A\in\mathcal{B}(F)$ and consider $$M=\{\langle Ax,x \rangle;\;x\in \text{Im}(A), \|x\|=1\}.$$
Let $c\in M$, then there exists $x\in \text{Im}(A)$ such that $\|x\|=1$ and $c=\langle Ax,x \rangle$.
If there exists $(x_n)_n\subset \text{Im}(A)$ such that $\|x_n-x\|\to 0$. Then we have $\langle Ax_n,x_n \rangle\to \langle Ax,x \rangle$. Further, clearly $\|x_n\|\to1$.
In this case it is true that $c=\langle Ax,x \rangle\in \overline{M}$? Where $\overline{M}$ is the closure of $M$ with respect to the topology induced by $\mathbb{C}$.
My problem is that we have only $\|x_n\|\to1$. However, I think in order to get $\langle Ax,x \rangle\in \overline{M}$, we should prove that $\|x_n\|=1$ for all $n$.
Note, that$$\lim_{n\to\infty}\frac{\langle Ax_n,x_n\rangle}{\|x_n\|^2}=\lim_{n\to\infty}\langle Ax_n,x_n\rangle=\langle Ax,x\rangle,$$since $\lim_{n\to\infty}\|x_n\|=1$. But then$$\lim_{n\to\infty}\left\langle A\frac{x_n}{\|x_n\|},\frac{x_n}{\|x_n\|}\right\rangle=\langle Ax,x\rangle.$$But each $\frac{x_n}{\|x_n\|}$ has norm $1$.