Q: Let $G:=C[-1,1]$ be the normed linear space of continuous real valued functions on the interval $[-1, 1]$ with the supremum norm. Let $K$ be the kernel of the linear functional $I:f\mapsto \int_{-1}^{1}f(x)dx$ on G. Consider the function $f (x) = x^2$ in G. What is its distance from the subspace $K$?
A: Let $g\in K$ be arbitrary. Then $\int_{-1}^{1}\|x^2-g(x)\|dx\geq\int_{-1}^{1}(x^2-g(x))dx$ gives $\|x^2-g(x)\|\geq1/3$. Also $x^2-1/3$ is at a distance $1/3$ from $x^2$ and happens to be a member of $K$. This shows $d(x^2,K)=inf_{g\in K}\|x^2-g(x)\|=1/3$.
Is this correct?
Your proof is written in a strange way so I had to ask for some aspects in the comment. Especially if you write $\|x^2-g(x)\|$ then it indicates you consider the norm in $\Bbb R$. Here is how it could look like imho:
Let $g\in K$ be arbitrary. Then $$2\cdot \|f-g\|=\int_{-1}^1\|f-g\|dx \geq \int_{-1}^1(x^2-g(x))dx = \int_{-1}^1x^2dx -0 = \frac 23.$$ This shows that $\|f-g\|\geq \frac 13$, so $\mathrm{dist}(f,K)\geq \frac 13$. On the other hand the function $g_0$ defined by $g_0(x)=x^2-1/3$ is a member of $K$, so $\mathrm{dist}(f,K)\leq \|f-g_0\| =\frac 13$.