0
$\begingroup$

I want to prove that the set $\{ f:||f||_{\infty}=1 \}$ where $f$ belongs to the space of continuous functions on $[a,b]$ is not a strictly convex set. As a counterexample, I'm asked to use $f(x)=x$ and $g(x)=x^2$ on $[0,1]$

I do understand the definition of convex set and strict convex set, but apparently not enough to prove this fact. I tried by contradiction, assuming it's convex in hope to find an element that is not an interior of the set. But I don't seem to get anywhere, so any help is appreciated. Thanks

$\endgroup$
  • $\begingroup$ Are you supposed to show that the set is not convex, or not strictly convex? $\endgroup$ – Umberto P. Feb 1 '18 at 21:02
  • $\begingroup$ @saulspatz which $t$ might that be? $\endgroup$ – Umberto P. Feb 1 '18 at 21:03
  • $\begingroup$ I think I'd rather show that all proper convex combinations of $f(x)=x$ and $g(x)=1-x$ are not in the set $\endgroup$ – Hagen von Eitzen Feb 1 '18 at 21:04
  • $\begingroup$ I would use $f(x)=x$, $g(x)=1-x$, and $\lambda=\frac{1}{2}$. $\endgroup$ – carmichael561 Feb 1 '18 at 21:04
  • $\begingroup$ Corrected it, prove that it's not STRICTLY convex $\endgroup$ – JustANoob Feb 1 '18 at 21:05
1
$\begingroup$

A set $S$ is called strictly convex if it is convex and furthermore all points $\lambda f + (1 - \lambda)g $ $\lambda \in (0,1)$ are in the interior of the set, for all $f,g \in S$.

The set can't be strictly convex because it is not even convex. As an example, pick $f(x)=x, g(x)=1-x$ on $[0,1]$ and $\lambda=1/2$. Clearly $f$ and $g$ belongs to the set. But $||\lambda g +(1-\lambda)f||_{\infty}= ||(1-x)/2+x/2||_{\infty}=1/2$, which shows that $\lambda g +(1-\lambda)f$ is not in the space, so can't be convex by definition.

$\endgroup$
  • $\begingroup$ Maybe he wants the full ball? $\endgroup$ – max_zorn Feb 1 '18 at 21:35
  • $\begingroup$ Edited the answer, this is the definition given by my professor. Is there different definitions? If so that's just confusing don't you think? $\endgroup$ – JustANoob Feb 1 '18 at 21:44
  • $\begingroup$ ok i get what you mean now. My professor had made a mistake in the question and it should be the whole unit ball... $\endgroup$ – JustANoob Feb 2 '18 at 14:28
0
$\begingroup$

Take $f = 1$ and $g = -1$ then $h = 0$ which is obviously the convex combination of $f$ and $g$ does not to belong the Unite sphere.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.