I am looking for a small clarification on why my teacher solved two similar problems in different ways. First, he wrote that $\langle f,g\rangle = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\overline{g(x)}\,dx$ and used this formula to solve the problem. However, in a later problem that was nearly identical, he did not multiply by $\frac{1}{2\pi}$. What would the correct formula be or why would you use both?
For context, here is the question: Define $T_{\cos x}(f): L^2(-\pi,\pi)\rightarrow L^2(-\pi,\pi)$ by $T_{\cos x}(f)=\langle f, \cos x\rangle \cos x$.
- What is $T_{\cos x}(f)$ when $f(x)=x^4$?
- Show $\lVert T_{\cos x}(x^4) \rVert \leq \lVert T_{\cos x} \rVert \cdot \lVert x^4 \rVert$
In the above questions my teacher did not divide the integral by $2\pi$, but in an earlier question where the only difference was $f(x)=x^2$, he did. What am I missing in my understanding?
He was using a different measure each time. $(f, g)_{L^2((-\pi, \pi), \,dx)} = \int_{-\pi}^{\pi} f(x)\overline{g(x)}\,dx$ is the inner product on $L^2((-\pi, \pi), \,dx)$, whereas $(f, g)_{L^2((-\pi, \pi), \frac{dx}{2\pi})} = \int_{-\pi}^{\pi}f(x)\overline{g(x)}\frac{dx}{2\pi}$ is the inner product on $L^2((-\pi, \pi), \frac{dx}{2\pi})$. Either way is fine, they only differ by a constant factor.