Periodic Functions & Orthogonal Functions — Question 2

Question 2 – Solution

Strategy. Distinguish exact simultaneous phase return from a uniformly small phase mismatch.

Step 1: Use the exact maximum. Each cosine is at most one. Equality $f(x)=2$ requires $x=2\pi m$ and $\sqrt 2\,x=2\pi n$ for integers $m,n$. Irrationality forces $m=n=0$, so the maximum is attained only at $x=0$. If $T>0$ were a period, then $f(T)=f(0)=2$, contradicting this uniqueness.

Step 2: Isolate the frequencies by differentiation. Direct differentiation gives $(D^2+2)f=\cos x$. Differentiating an identity $f(x+T)=f(x)$ twice preserves it, so $T$ would also be a period of $\cos x$. Subtracting this cosine from $f$ shows that $T$ preserves $\cos(\sqrt 2x)$ too. Evaluating both shifted cosines at zero requires $T=2\pi m$ and $\sqrt 2T=2\pi n$, impossible for positive $T$.

Step 3: Bound an approximate return. At $T=140\pi$, the first cosine returns exactly. The second phase shift is $2\pi(70\sqrt 2)=2\pi\cdot 99-\delta$, where $$\delta=2\pi(99-70\sqrt 2)=\frac{2\pi}{99+70\sqrt 2}.$$ The stated cosine inequality therefore yields $$\boxed{\sup_{x\in\mathbb R}|f(x+140\pi)-f(x)|\leq \frac{2\pi}{99+70\sqrt 2}<0.032.}$$

Step 4: Contrast approximate and exact recurrence. The discrepancy bound is positive and does not make $T$ a period; Step 1 rules out every positive exact period. For $h$, any period must send its value two at zero to another value two. Thus $T=2\pi m$ and $3T/2=2\pi n$, forcing $m$ even. Conversely $4\pi$ preserves both summands. Hence $\boxed{T_h=4\pi}$, an exact fundamental period.