Periodic Functions & Orthogonal Functions — Question 9

Question 9 – Solution

Strategy. Orthogonal measurements determine a finite projection, while all perpendicular components remain unseen.

Step 1: Recover the measured component. The squared norms of $1,\cos x,\sin x$ are $2\pi,\pi,\pi$. Dividing each moment by its corresponding squared norm gives $$\boxed{p(x)=1+\cos x-2\sin x,\qquad\|p\|^2=7\pi.}$$ Orthogonality makes these coefficients unique within the three-dimensional span.

Step 2: Derive the sharp energy bound. Every compatible $f$ has the form $\boxed{f=p+r}$, where $r$ is continuous and orthogonal to all three measurement functions. Conversely every such residual preserves the measurements. Since $\langle p,r\rangle=0$, $$\boxed{\|f\|^2=7\pi+\|r\|^2\geq 7\pi.}$$ Equality holds exactly when $r=0$ everywhere: a continuous function with zero squared integral is identically zero. Thus $p$ is the unique minimum-energy signal.

Step 3: Exhibit the invisible freedom. Product-to-sum identities give zero inner products between $\cos 2x$ and each of $1,\cos x,\sin x$. Therefore $$\boxed{f_C=p+C\cos 2x,\quad C\in\mathbb R,\qquad\|f_C\|^2=7\pi+C^2\pi.}$$ The measurement functions are independent, but their span does not contain all continuous functions. The data determine the projection, not its orthogonal complement.

Step 4: Test the fourth measurement. The added measurement removes the particular $\cos 2x$ freedom above, but $\boxed{p+D\sin 2x}$ satisfies all four conditions for every real $D$. The required product integrals again vanish by parity or product-to-sum. A finite orthogonal family supplies independent coordinates for its own span; it is not a complete representation of all continuous signals on the interval.