Norms for $L^2$?

Question

This is a basic question, but it kind of is confusing me. I know that the inner product in $L^2[-\pi,\pi]$ is given by $$\langle f,g\rangle=\int_{-\pi}^\pi f(x)\overline{g(x)}\,dx$$ Then we can get an norm that is defined by $||f||=\sqrt{\langle f,f\rangle}$. But then when I was also reading that for an $L^p$ space, the norm is usually defined as $$||f||=\left(\int_{-\pi}^\pi|f(x)|^p\,dx\right)^{\frac{1}{p}}$$ Are these norms equal when $p=2$? And if not, if I were to show that $L^2$ is complete, would I just go with the first norm?

Answer 1

These norms are equal when $p = 2$, because $$\langle f, f \rangle = \int_{-\pi}^\pi f(x) \overline{f(x)} dx = \int_{-\pi}^\pi |f(x)|^2 dx = \left(\int_{-\pi}^\pi |f(x)|^2 dx \right) ^{2 \times 1/2} = ||f||_2^2.$$

In the case of any $p\neq 2$, one can show that the $L^p$-norm does not arise from an inner product, so $p=2$ is the only case where something like this happens.

If you want to show that $L^2$ is complete yourself, the extra structure of an inner product might be helpful. But $L^p([-\pi, \pi])$ is complete for $1 \leq p \leq \infty$ (this is the Riesz-Fisher theorem), and standard proofs of this handle all cases except for $p = \infty$ at once, so one does not need to think of $||f||_2$ coming from an inner product to prove this.

Answer 2

Yes, this is the same norm $$ \sqrt{\langle f,f\rangle} = \left(\int_{-\pi}^{\pi} f(x)\,\overline{f(x)}\,\mathrm{d}x\right)^{\frac{1}{2}} = \left(\int_{-\pi}^{\pi} |f(x)|^{2}\,\mathrm{d}x\right)^\frac{1}{2}. $$ since $z\bar{z} = |z|^2$. Was just this your question?