= Solution
For $\alpha>0$, minimize $J_\alpha(x)=\|Ax-y\|^2+\alpha\|x\|^2$. Its first variation in every direction gives
$$
(A^*A+\alpha I)x_\alpha=A^*y.
$$
The <self-adjoint operator> on the left has lower bound $\alpha I$, so it is invertible and the minimizer is unique. The regularized solution has the singular expansion
$$
\boxed{x_\alpha=\sum_j\frac{\sigma_j}{\sigma_j^2+\alpha}(y,v_j)u_j.}
$$
Since $\sigma^2+\alpha\ge2\sigma\sqrt\alpha$, its gain is at most $1/(2\sqrt\alpha)$. Consequently
$$
\boxed{\|x_\alpha(y^\delta)-x_\alpha(y)\|\le\frac{\|y^\delta-y\|}{2\sqrt\alpha}.}
$$
This proves existence, uniqueness and continuous data dependence at fixed positive parameter. For the particular perturbation from part (i), the error is $\delta\sigma_j/(\sigma_j^2+\alpha)$ instead of $\delta/\sigma_j$. To recover a minimum-norm exact solution as noise tends to zero, choose $\alpha\to0$ while $\delta/\sqrt\alpha\to0$; <dominated convergence> of the exact-data singular coefficients removes the bias.
To connect this with the provided inverse expansion, put $B=AA^*$ and $C=A^*A$. The exact intertwining identity is
$$
(A^*A+\alpha I)^{-1}A^*=A^*(\alpha I+AA^*)^{-1}.
$$
It follows by multiplying $(C+\alpha I)A^*=A^*(B+\alpha I)$ on either side by the appropriate <resolvent>. A useful exact inverse identity is
$$
(\alpha I+B)^{-1}=\alpha^{-1}I-\alpha^{-2}A(I+\alpha^{-1}C)^{-1}A^*.
$$
Expanding $(I+\alpha^{-1}C)^{-1}$ to two terms gives the printed approximation, namely
$$
R_{\alpha,2}=\alpha^{-1}I-\alpha^{-2}B+\alpha^{-3}B^2.
$$
This is the <truncated Tikhonov resolvent>. Multiplication gives $(\alpha I+B)R_{\alpha,2}=I+\alpha^{-3}B^3$, so the expression is not an exact inverse. For example $A=1$ and $\alpha=1$ give the approximate value $1$ instead of the exact $1/2$. Its exact remainder is
$$
R_{\alpha,2}-(\alpha I+B)^{-1}=\alpha^{-3}B^3(\alpha I+B)^{-1},\qquad
\|R_{\alpha,2}-(\alpha I+B)^{-1}\|\le\frac{\|B\|^3}{\alpha^4}.
$$
In the regime $\|C\|/\alpha<1$ it is a controlled Neumann expansion. Using it to reconstruct gives
$$
x_{\alpha,2}=\left(\alpha^{-1}I-\alpha^{-2}C+\alpha^{-3}C^2\right)A^*y.
$$
For each fixed $\alpha>0$ this is a bounded <polynomial> in bounded operators, with
$$
\|x_{\alpha,2}(y^\delta)-x_{\alpha,2}(y)\|
\le\left(\frac{\|A\|}{\alpha}+\frac{\|A\|^3}{\alpha^2}+\frac{\|A\|^5}{\alpha^3}\right)\|y^\delta-y\|.
$$
Thus the supplied approximation also has continuous data dependence in its domain of use. It must not be extrapolated to $\alpha\downarrow0$ with $A$ fixed: its gain on any nonzero singular mode diverges, so it would not define a convergent regularization family there. \b[The exact Tikhonov filter supplies the well-posed regularized problem; the truncated formula is only a controlled computational approximation.]
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