4
$\begingroup$

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 $$||y_0-x_0||=\operatorname{dist}(y_0,V)=\inf_{x\in V}||y_0-x||$$

$\endgroup$
1
  • 1
    $\begingroup$ 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? $\endgroup$ Commented Sep 2, 2013 at 5:12

1 Answer 1

11
$\begingroup$

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, $f(t)=t$.

$\endgroup$
1
  • $\begingroup$ Could you please a hint of how to prove the first fact? The one where you expressed the norm as a function of the distance of the function to its Kernel. $\endgroup$
    – user770533
    Commented Nov 19, 2021 at 15:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .