Consider $H$ be a separable Hilbertian space with real inner product $\langle , \rangle $. Let $\Gamma: H \to H$ be a Nuclear operator $$\Gamma(u) = \sum_{n=1}^\infty \lambda_n \langle u, \phi_n\rangle \phi_n$$ And it's squared root (which is a compact operator) $$\Gamma^{1/2}(u) = \sum_{n=1}^\infty \lambda_n^{1/2} \langle u, \phi_n\rangle \phi_n$$ Where $(\phi_n)_{n \in \mathbb{N}}$ form an orthonormal basis for H. And $\lambda_1 \geq \lambda_2 \geq \ldots > 0$ such that $\sum_{n=1}^\infty \lambda_n < \infty$ Now let's define the norm associated to $\Gamma$ $$ \| u \|^2_{\star} = \| \Gamma^{1/2} (u) \|^2 = \sum_{i=1}^\infty \lambda_i \langle u, \phi_i\rangle^2 $$
Does weak convergence implies convergence in the new norm ?
i.e $$\langle u_n, v \rangle \to \langle u , v\rangle \quad \forall v \in H \implies \| u_n - u \|_\star \to 0 $$
It is easy to see the converse does not hold because $\| \lambda_n^{-1/2} \phi_n \|_{\star}^2 = \lambda_n^{1/2} \to 0$ but $\lambda_n^{-1/2} \phi_n$ is unbounded which contradicts the uniform boundness principle.
Let $u_n\rightharpoonup u$. Fix some index $N$. Then $$ \|u_n-u\|_*^2 = \sum_{i=1}^\infty \lambda_i\langle u-u_n,\phi_i\rangle^2 = \sum_{i=1}^N \lambda_i\langle u-u_n,\phi_i\rangle^2 + \sum_{i=N+1}^\infty \lambda_i\langle u-u_n,\phi_i\rangle^2\\ \le \sum_{i=1}^N \lambda_i\langle u-u_n,\phi_i\rangle^2 + \lambda_{N+1}\|u-u_n\|^2, $$ where the inequality follows from the monotonicity of $(\lambda_i)$ and the Parseval identity. Now choose $N$ large enough such that the second addend becomes small. Then the first addend can be made small for $n$ large enough.