Why is $\|f-s_n(f)\|_2=\inf_{T\in\mathcal{T}_n}\|f-T\|_2$

30 Views Asked by At

I am working through some examples in my book in the section on Fourier Series.

Why is $\|f-s_n(f)\|_2=\inf_{T\in\mathcal{T}_n}\|f-T\|_2$?

where $f$ is a continuous $2\pi$ periodic function, $T$ is some trig. polynomial, and $s_n(f)$ is a trig. polynomial of degree at most $n$, $\mathcal{T}_n$ is the space of all trig. polynomials of degree $n$

Thoughts:

Writing this out, I get:

$$\left(\dfrac{1}{\pi}\int_{-\pi}^{\pi}(f-s_n(f))^2\right)^{1/2}=\inf_{T\in\mathcal{T_n}}\left(\dfrac{1}{\pi}\int_{-\pi}^{\pi}(f-T)^2\right)^{1/2}$$

I understand that both $s_n(f)$ and $T$ are both trig. polynomials of degree at most $n$, but I'm not seeing why this means that the two integrals are necessarily equal.

Any guidance would be appreciated. Thanks.