I'd like to prove that: $$ \|u_\varepsilon-f\|_*\rightarrow0 \quad\text{in }V^* $$ with $V$ Hilbert and $V^*$ its dual. In particular $u_\varepsilon\in V$. From the precedent points of the proof I have that $\{w_j\}$ is a orthogonal basis of $V$ (obtained with the sprectral theory and $\{\lambda_n\}$ are eigenvalues of the bilinear form of the exercise) and we saw that can be used also for the element of the dual. So I have that $$ u_\varepsilon=\sum_{j=1}^\infty \frac{f_j}{1+\varepsilon^2\lambda _j}w_j\qquad f=\sum_{j=1}^\infty f_jw_j $$ (this from the precedent points of the proof) Becuase $V$ has $\dim V=\infty$ the convergence of the coordinates isn't enough, right? So now I have to work with $\|\cdot\|_*$.
$$ \left\|\sum_{j=1}^\infty \frac{f_j}{1+\varepsilon^2\lambda _j}w_j- \sum_{j=1}^\infty f_j w_j \right\|_*=\left\| \sum_{j=1}^\infty \left( \frac{\varepsilon^2\lambda_j}{1+\varepsilon^2 \lambda_j}\right)f_jw_j\right\|_* $$
At this point I thought to work in $V_m=\operatorname{span}\{w_1,\ldots,w_m\}$ in order to work with a finite number of elements in the sums and use without problem the following passage: $$ \left\| \sum_{j=1}^m \left( \frac{\varepsilon^2\lambda_j}{1+\varepsilon^2 \lambda_j}\right)f_jw_j\right\|_*=\sum_{j=1}^m\left\| \left( \frac{\varepsilon^2\lambda_j}{1+\varepsilon^2 \lambda_j}\right)f_jw_j\right\|_* $$ At this point can I do: $\varepsilon\rightarrow 0$ and then work with $m\rightarrow \infty$? Obviously I'll have to discuss the convergence of the series in order to do the limit for $m\rightarrow \infty$. I tried to do the best in order to describe my problem. If something isn't clear I'll give your any necessary information that I missed.
I do not understand why you use the dual, Hilbert spaces being self-dual. Secondly I think that your last formula, where you bring the summation outside the norm, is incorrect.
Since the $w_{j}$ 's are orthonormal and assuming the $\lambda _{j}$ 's are real (complex is OK but the notation is more involved) \begin{eqnarray*} \left\Vert \sum_{j=1}^{\infty }\frac{\varepsilon ^{2}\lambda _{j}}{% 1+\varepsilon ^{2}\lambda _{j}}f_{j}w_{j}\right\Vert ^{2} &=&\left( \sum_{j=1}^{\infty }\frac{\varepsilon ^{2}\lambda _{j}}{1+\varepsilon ^{2}\lambda _{j}}f_{j}w_{j},\sum_{h=1}^{\infty }\frac{\varepsilon ^{2}\lambda _{h}}{1+\varepsilon ^{2}\lambda _{h}}f_{h}w_{h}\right) \\ &=&\sum_{j=1}^{\infty }\left( \frac{\varepsilon ^{2}\lambda _{j}}{% 1+\varepsilon ^{2}\lambda _{j}}\right) ^{2}|f_{j}|^{2} \end{eqnarray*} Now you can use that $$ \left( \frac{\varepsilon ^{2}\lambda _{j}}{1+\varepsilon ^{2}\lambda _{j}}% \right) ^{2}\leqslant 1 $$ and $$ \sum_{j=1}^{\infty }|f_{j}|^{2}<\infty $$ to restrict the summation making an arbitrarily small error. Then, in each of the finite set of remaining terms $$ \frac{\varepsilon ^{2}\lambda _{j}}{1+\varepsilon ^{2}\lambda _{j}}% \rightarrow 0 $$