A counter example of best approximation

Construct a point $f\in C[0,1]$ and a closed subspace $V\subset C[0,1]$ such that $f$ does not have a best approximation in $V$.

Definition: $C[0,1]$ is the set of countinous function with the norm $||f||=\max\limits_{x\in[a,b]}|f(x)|$.

Let $B$ be a normed vector space with norm $||\cdot||$. Set $V\subset B$ and $y_0\in B$. A vector $x_0\in V$ is called the best approximation of $y_0$ in V if $$||y_0-x_0||=\operatorname{dist}(y_0,V)=\inf_{x\in V}||y_0-x||$$

• Welcome to Math SE! Will you tell us a bit of where you've encountered the question (if this is homework it should be labeled as such, for example) and what approaches you've tried so far? Commented Sep 2, 2013 at 5:12

One can use the following general fact (whose proof is not difficult): if $\Phi$ is any continuous linear functional on $B$ and if $f\in B\setminus \ker(\Phi)$, then $$\Vert\Phi\Vert=\frac{\vert\Phi(f)\vert}{{\rm dist}(f,\ker(\Phi))}\, .$$ It follows that if you start with any linear functional $\Phi$ which does not attain its norm (i.e. there is no $x$ with $\Vert x\Vert=1$ and $\Phi(x)=\Vert\Phi\Vert$), then $V=\ker\Phi$ and any $f$ such that $\Phi(f)\neq 0$ do the job.
In the special case you are looking at, $V=\mathcal C([0,1])$, consider the linear functional $\Phi$ defined by $$\Phi(x)=\int_0^{1/2} x(t)\, dt-\int_{1/2}^1 x(t)\, dt\, .$$ It is not hard to check that $\Vert\Phi\Vert=1$ and that $\Phi$ does not attain its norm (the idea is that if $x\in\mathcal C([0,1])$ satisfied $\Vert x\Vert=1=\Phi(x)$, then $x$ would have to be equal to $1$ on $(0,1/2)$ and to $-1$ on $(1/2,1)$, so $x$ would not be in the space $\mathcal C([0,1])$.)
In conclusion, you may take $$V=\left\{ x\in\mathcal C([0,1]);\; \int_0^{1/2}x(t)\, dt=\int_{1/2}^1x(t)\, dt \right\}$$ and, for example, $f(t)=t$.