Series convergence in Hilbert space and dual. 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.
 A: 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
$$
\frac{\varepsilon ^{2}\lambda _{j}}{1+\varepsilon ^{2}\lambda _{j}}%
\rightarrow 0
$$
