I am currently self-studying Linear algebra done right and have reached the part about the minimization problem
$6.58 $Example Find a polynomial u with real coefficients and degree at most $5$ that approximates $\sin(x)$as well as possible on the interval $[-\pi,\pi]$ in the sense that $$\int_{-\pi}^{\pi}|\operatorname{u}(x)-\sin(x)|^2dx$$ is as small as possible. Compare this result to the Taylor series approximation. Solution Let $\mathbb C_{\mathbb R}[-\pi, \pi]$ denote the real inner product space of continuous real-valued functions on $[-\pi, \pi]$ with inner product $$ \langle f,g\rangle = \int_{-\pi}^{\pi}f(x)g(x)dx$$ $6.59$ Let $v \in \mathbb C_{\mathbb R}[-\pi, \pi]$ be the function defined by $v(x)=\sin(x)$ Let $U$ denote the subspace of $\mathbb C_{\mathbb R}[-\pi, \pi]$ consisting of the polynomials with real coefficients and degree at most $5$. Our problem can now be reformulated as follows: Find $u \in U$ such that $\Vert u-v \Vert$ is as small as possible. To compute the solution to our approximation problem, first apply the Gram–Schmidt Procedure (using the inner product given by $6.59$) to the basis $1, x, x^2, x^3, x^4, x^5$ of $U$, producing an orthonormal basis $e_{1}, e_{2}, e_{3}, e_{4}, e_{5}$ of $U$. Then, again using the inner product given by 6.59, compute $P_{U}v$ using $6.55$(i) (with $m=6$). Doing this computation shows that $P_Uv$ is the function $u$ defined by $$6.60 \space\space\space\space\space\space u(x) = 0.987862x - 0.155271x^3 + 0.00564312x^5$$ where the $\pi$’s that appear in the exact answer have been replaced with a good decimal approximation.
My question is, in the derivation of the orthogonal complement, Mr. Axler uses the fact that the inner product space V is finite dimensional. However, here $\mathbb C_{\mathbb R}[-\pi, \pi]$ is an infinite dimensional inner product space. Does this mean that the proof does not apply here?
$C_{\mathbb R}([-\pi, \pi])$ is indeed infinite-dimensional, but here all the action takes place in the finite-dimensional subspace spanned by $U$ and $v$. So we might as well use that as our inner product space. All results for finite-dimensional inner product spaces are then fair game.