Let \(Q\) be a bounded linear operator and set \(e=j-\widehat j\). Substituting \(f=s+j\) into \(\widehat f=Q(f-\widehat j)+\widehat j\) gives
The second term includes the action of \(Q\) on the packet error. Linearity is required for this split.
Reading the drawing
Two vectors show the packet discrepancy and its image under the complementary projection. Labels give the corresponding exact Euclidean norms.
Drawing labels and data 4 labels
Read the original labels as selectable text. Tables preserve grid relationships; other short labels are grouped compactly.
- \(e=(0,1)\)
- \((I-Q)e=(-2,1)\)
- \(-2\)
\(\|e\|=1\)
[5pt]\(\|(I-Q)e\|=\sqrt5\)
An oblique projection can increase the error
In \({\mathbb R}^2\), choose
Then \(Q^2=Q\) and \(Qs=s\), but \((I-Q)e=(-2,1)\). The smooth error is zero while the unit packet error becomes a reconstruction error of norm \(\sqrt5\). For a fully specified realization, take \(j=e\) and \(\widehat j=0\). Then \(f=(1,1)\), \(\widehat f=Qf=(3,0)\), and their difference is exactly \((-2,1)\).
Use the actual operator bound
The always-valid estimate in this setting is
Here \(\|Q\|=\sqrt5\), so the bound is conservative. For an orthogonal projection, \(\|I-Q\|\le1\) gives a sharper estimate. The operator's geometry determines which bound can be used.
Read this as
Recovering the packet and approximating the background are separate tasks, but the chosen approximation operator couples their final errors.