What is the norm of the vector \(f(x) = \cos(x)\) in the inner product space \(C([0, 2\pi])\)? The inner product is defined by \(\langle f(x), g(x)\rangle = \int_0^{2\pi} f(x).g(x)\,dx\)
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Compute the inner product of cos(x) with itself over [0, 2*pi] using the half-angle identity, then take the square root.
Step 1: Norm formula. The norm induced by an inner product is $\|f\| = \sqrt{\langle f,f\rangle}$, so the problem reduces to computing $\langle \cos x, \cos x\rangle = \int_0^{2\pi}\cos^2x\,dx$ and taking a square root at the end.
Step 2: Use the average-value trick over a full period instead of the half-angle expansion. Over one complete period $[0,2\pi]$, $\cos^2 x$ oscillates symmetrically between 0 and 1, and its average value over a full period is exactly $\frac{1}{2}$ (a standard fact since $\cos^2x$ and $\sin^2x$ have equal average value and together average to 1). So \[ \text{average of } \cos^2 x \text{ over } [0,2\pi] = \frac{1}{2} \]
Step 3: Convert the average value to the definite integral. \[ \int_0^{2\pi}\cos^2 x\,dx = (\text{average value}) \times (\text{length of interval}) = \frac{1}{2}\times 2\pi = \pi \] This agrees with the value obtained from direct integration using $\cos^2x = \frac{1+\cos2x}{2}$.
Step 4: Take the square root. \[ \|f\| = \sqrt{\pi} \]
Step 5: Quick elimination check. $0$ is impossible since $\langle f,f\rangle$ is strictly positive. $\sqrt{2\pi}$ would require an average value of 1, meaning $\cos^2 x \equiv 1$ everywhere, false. $\sqrt{\pi/2}$ would require an average value of $\frac{1}{4}$, which does not match the known $\frac12$ average of $\cos^2x$. \[ \boxed{\|f\| = \sqrt{\pi}} \]
