= Solution
Using the <singular value system> with $Au_j=\sigma_jv_j$ and $A^*v_j=\sigma_ju_j$, the formal minimum-norm inverse is
$$
x=\sum_j\frac{(y,v_j)}{\sigma_j}u_j.
$$
It defines an element of the unknown <Hilbert space> only if $\sum_j|(y,v_j)|^2/\sigma_j^2<\infty$, the <Picard criterion>. A null-space component is undetermined by the data.
Adding an error along a single <left singular vector> changes precisely one coefficient:
$$
x_\delta-x=\frac\delta{\sigma_j}u_j,\qquad
\boxed{\|x_\delta-x\|=\frac\delta{\sigma_j}.}
$$
Although this perturbation has data <norm> $\delta$, its amplification factor is $1/\sigma_j$. If $\sigma_j\to0$, these factors are unbounded. More explicitly choose $y^{(j)}=\sigma_jv_j$, tending to zero, while the minimum-norm inverse is $u_j$, of <norm> one. Thus the inverse is discontinuous at zero. It may also fail to exist for data outside its nonclosed range, so merely minimizing a residual does not guarantee that the formal inverse exists. This is \b[<ill-posedness> by unbounded small-singular-value amplification], not a claim that every finite-dimensional or bounded forward map has an unstable inverse.
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