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Suppose we have a linear PDE like $-\Delta u + \lambda u = 0$ with appropriate boundary conditions, so that there is a unique solution. One way to prove wellposedness is using Galerkin approximations. Here we have a sequence $u_n$ solving some regularised problem and we obtain $\lVert u_n\rVert \leq C$ in some Hilbert space (let's say) $H$ with $C$ independent of $n$.

Thus there is a weakly convergent subsequence $u_{n_j}$ to a limit point $u_1$. But there may be more than one (eg. consider $u_n = (-1)^n$) weakly convergent subsequence, so suppose we have two limit points $u_1$ and $u_2$ that are limits of two different subsequences.

Then passing to the limit we find that both $u_1$ and $u_2$ solve the PDE with $u_1 \neq u_2$, but this cannot be since we have uniqueness. So what's the fault in this argument?

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There is no fault in the argument. You have to proof, that $u_1 \equiv u_2$. And that will only be true, as you said, if you use an appropriate set of conditions. So: By solving the PDE you get a bunch of functions, you get a unique solution by applying boundary conditions. Boundedness of $u_n$ alone won't get you anywhere.

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You can apply the convergence principle:

If every subsequence contains a subsequence converging weakly to $u$, then the whole sequence converges.

In your problem, you take a weakly converging subsequence. Now the weak limit satisfies the equation, hence it is equal to the solution of the equation. Now apply the principle above.

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