Question

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 \(\langle f(x), g(x)\rangle = \int_0^{2\pi} f(x).g(x)\,dx\)

• A. \( \sqrt{\pi} \)
• B. \( \sqrt{2\pi} \)
• C. 0
• D. \( \sqrt{\pi/2} \)

Hint:

Compute the inner product of cos(x) with itself over [0, 2*pi] using the half-angle identity, then take the square root.

Answer 1

The Correct Option is A

Step 1: Recall the definition of a norm from an inner product.
In any inner product space, the norm of a vector \(f\) is defined as the square root of the inner product of the vector with itself: \[ \|f\| = \sqrt{\langle f, f \rangle} \]

Step 2: Set up the required integral.
Here \(f(x) = \cos(x)\), and the given inner product is \(\langle f, g \rangle = \int_0^{2\pi} f(x)\,g(x)\,dx\). So \[ \langle f, f \rangle = \int_0^{2\pi} \cos(x)\cos(x)\,dx = \int_0^{2\pi} \cos^2(x)\,dx \]

Step 3: Evaluate the integral using the half-angle identity.
Using \(\cos^2(x) = \dfrac{1 + \cos(2x)}{2}\), \[ \int_0^{2\pi} \cos^2(x)\,dx = \frac{1}{2}\int_0^{2\pi} 1\,dx + \frac{1}{2}\int_0^{2\pi} \cos(2x)\,dx \] The first term gives \(\frac{1}{2}(2\pi) = \pi\). The second term integrates to \(0\) because \(\sin(4\pi) - \sin(0) = 0\). So \[ \langle f, f \rangle = \pi \]

Step 4: Take the square root to get the norm.
\[ \|f\| = \sqrt{\langle f, f \rangle} = \sqrt{\pi} \]

Step 5: Rule out the other options.
\(\sqrt{2\pi}\) would result if the factor of 1/2 in \(\cos^2 x\) were omitted. \(0\) is wrong because \(\langle f, f\rangle\) is a sum of squares and cannot vanish since \(\cos x\) is not identically zero. \(\sqrt{\pi/2}\) would arise from mistakenly integrating over only half the interval. \[ \boxed{\|f\| = \sqrt{\pi}} \]