Consider $C[-1,1]$, the normed linear space of real-valued continuous functions on $[-1,1]$ with the supremum norm. Let K be the kernel of the linear functional $I: f \to \int_{-1}^{1}f(x)dx $ on $C[-1,1]$. What is the distance of the function $f(x)=x^2$ from $K$?
Clearly, $f(x)=x^2$ is not in the kernel $K$, and so the distance has to be non-zero. Also, the orthogonal projection formula doesn't seem to work here because the kernel is infinite dimensional. I am stumped. How do I even begin to attack this problem?
Indeed, @StiftungWarentest is right, the distance is $d(f,K) = \frac13$ and $g(x) = x^2-\frac{1}{3}$ is the distance minimizer.
First we can find the distance minimizer in the $\|\cdot\|_2$-norm. Namely, then $K^\perp$ is the space of constant functions whose orthonormal basis is $\left\{\frac1{\sqrt{2}}\right\}$ so the orthogonal projection of $f$ onto $K$ is given by $$P_K f = (I-P_{K^\perp })f = f - \left\langle f, \frac1{\sqrt{2}}\right\rangle \frac1{\sqrt{2}} = f - \frac12 \int_{-1}^1 f(x)\,dx$$ which indeed turns out to be equal to $g$. Since $\|\cdot\|_2 \le \sqrt{2}\|\cdot\|_\infty$ for all $h\in K$ we have $$\frac{\sqrt{2}}{3} = \left\|\frac13\right\|_2=\|f-g\|_2 \le \|f-h\|_2 \le \sqrt{2}\|f-h\|_\infty$$ and therefore $\|f-h\|_\infty \ge \frac13$. On the other hand, we have $\|f-g\|_\infty = \left\|\frac13\right\|_\infty = \frac13$ so we conclude that $d(f,K) = \frac13$.