Let $C[-1,1]$ and $f:[-1,1]\to \mathbb{C}$ with the inner product $\langle f,g\rangle=\int_{-1}^{1}f(x)\overline{g(x)}dx$

Prove: $P_{0}=1, P_{1}=x, P_{2}=1-3x^2$ is orthogonal in $C[-1,1]$

All $\langle P_{0},P_{1}\rangle,\langle P_{0},P_{2}\rangle,\langle P_{2},P_{1}\rangle$ are equal to zero, so they are orthogonal to each other, in the book that took the norm of each $P_{i}$ to show it is not zero, why should we do it? is it to check that all $P_{i}\neq 0$ because if they were it is "trivial" orthogonal system?


1 Answer 1


Actually, it all depends on what you admit first.

Recall that an inner product must verify $\langle v,v \rangle = 0 \Leftrightarrow v = 0$ for all $v$ in the vector space.

So, if you admit the application you defined here is an inner product, then you know that for any function $f$ different from $0$, you will have $\langle f,f \rangle \not = 0$, and you do not need to verify it.

I find it quite weird that the book checks this afterwards. If it was a problem, the exercise should have asked you to show $\int_{-1}^1 f(x)\overline{g(x)}\mathrm{d}x$ defines an inner product. Otherwise, asking you to prove $P_0, P_1$ and $P_2$ are orthogonal make no sense, you need an inner product to give sense to the orthogonality (actually, you don't really, but I don't think going into quadratic forms is the problem here).


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