Suppose one has:
$$\vartheta:=\|f(E)\|<\infty$$
Regard the sets:
$$\Omega_n:=\{\vartheta+\frac{1}{n}\leq|f|\leq\vartheta+n\}\in\mathcal{B}(\mathbb{C})$$
And their union:
$$\Omega:=\bigcup_{n=1}^\infty\Omega_n=\{|f|>\vartheta\}\in\mathcal{B}(\mathbb{C})$$
Regard the functions:
$$1_n:=\chi_{\Omega_n}:\quad 1_n(E)=E(\Omega_n)$$
Denote for readability:
$$f_n:=f1_n:\quad\|f_n\|_\infty<\infty$$
For them one has:
$$\mathcal{D}f_n(E)=\mathcal{H}:\quad f_n(E)=f(E)1_n(E)$$
So one obtains inequality:
$$\left(\vartheta+\tfrac{1}{n}\right)^2\|E(\Omega_n)\varphi\|^2=\int\left(\vartheta+\tfrac{1}{n}\right)^21_n^2\mathrm{d}\nu_\varphi\\
\leq\int|f|^21_n^2\mathrm{d}\nu_\varphi=\|f_n(E)\varphi\|^2=\|f(E)1_n(E)\varphi\|^2\leq\vartheta^2\|E(\Omega_n)\varphi\|^2$$
That only holds for:
$$E(\Omega_n)=0\implies E(\Omega)=0$$
So it was bounded:
$$E\{|f|>\vartheta\}=E(\Omega)=0$$
Conversely one has:
$$\|f(E)\varphi\|^2=\int|f|^2\mathrm{d}\nu_\varphi\leq\|f\|_\infty\|\varphi\|<\infty$$
Therefore it holds:
$$\mathcal{D}f(E)=\mathcal{H}:\quad\|f(E)\|\leq\|f\|_\infty\leq\|f(E)\|$$
Concluding the assertion.
