1
$\begingroup$

I want to prove that the set $\{ f:\lVert f\rVert_{\infty}\leq 1 \}$ 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 have posted a question before for precisely the same question but for the set $\{ f:\lVert f\rVert_{\infty} = 1 \}$. It turns out that the set is not even convex, so that was easy(after some help).

But for this one i just cant figure it out. The set is convex indeed, but for strict convexity I am a bit lost. My approach was to show that there exists no $\epsilon >0$ such that the ball $B(\lambda x + (1 -\lambda)x^2, \epsilon)$ is contained in the set, for $\lambda \in (0,1)$. But I don't seem to get anywhere, so any help is appreciated. Thanks

$\endgroup$
  • $\begingroup$ What is $\|f\|_{\infty}$? What is $\|g\|_{\infty}$? What is $\|\lambda f + (1 - \lambda) g\|_{\infty}$ if $0 < \lambda < 1$? $\endgroup$ – Hans Engler Feb 2 '18 at 14:47
  • $\begingroup$ Check that the line between $x$ and $x^2$ lies on the boundary of your convex set. $\endgroup$ – Jeff Feb 2 '18 at 14:47
  • $\begingroup$ So strict convexity means that for any two points $f,g\in M$ their connecting line lies within the interior of $M$, with the possible exception of the endpoints $f,g$ themselves. Intuitively it should make sense that the boundary of $M = \{f: \|f\|\le 1\}$ is $\{f:\|f\|=1\}$. So try to find a $\lambda\in (0,1)$ such that $1 = \|\lambda x + (1-\lambda)x^2\|$, which would be some point in between $x,x^2$ that also lies on the boundary. $\endgroup$ – Hyperplane Feb 2 '18 at 14:52
  • $\begingroup$ $\lambda $= 1/2 does the job. Intuitively it should be as you say, i just don't think i can answer in that way. $\endgroup$ – JustANoob Feb 2 '18 at 15:01
1
$\begingroup$

The definition of strict convexity in topological vector spaces is the following: $\Omega$ is strictly convex if it is convex and for all $x,y\in \partial \Omega$, $x\neq y$, one have $(1-t)x+t y\in int(\Omega)$, $t \in (0,1)$. First we prove that it is convex: we take any two functions $f,g\in \Omega$, then $$ \|(1-t)f+tg\|_\infty \leq (1-t) \|f\|_\infty + t \|g\|_\infty \leq 1 \quad \Rightarrow \quad (1-t)f+tg \in \Omega. $$ Now I claim that $\|f\|=1$ ($\|\cdot\|=\|\cdot\|_\infty)$ if and only if $f\in\partial \Omega$. Suppose that $f\in\partial\Omega$ but $\varepsilon = 1-\|f\|>0$, so that for all $h\in B_\varepsilon(0)$ (i.e. $\|h\|\leq \varepsilon$), we get $$ \|f+h\|\leq \|f\|+\|h\| = 1-\varepsilon + \|h\| \leq 1-\varepsilon + \varepsilon = 1, $$ which means $f + B_\varepsilon(0) = B_\varepsilon(f)\subset \Omega$, contradicting the fact that $f\in\partial \Omega$. Viceversa, if $\|f\|=1$ then by usual compactness argument there is $x\in [a,b]$ such that $|f(x)|=1$. W.l.g assume $f(x)=1$. For all $1>\varepsilon>0$, $q=\varepsilon$, then $f+q \not\in \Omega$. Moreover, for $-\varepsilon f$, we get $$ f -\varepsilon f = f(1-\varepsilon), \quad \|f-\varepsilon f\| \leq (1-\varepsilon) \in \Omega.$$ This imply that, for all $\varepsilon>0$, $B_\varepsilon(f)$ intersects both $\Omega \setminus \{f\}$ and $\Omega^c$, i.e. is a boundary point

Now $\|x\|=1 =\|x^2\|$, so that $x,x^2\in\partial\Omega$, although $$ 1\geq \|tx+(1-t)x^2\|\geq t(1)+(1-t)(1)^2 = t-1+t = 1, $$ that is $\|tx+(1-t)x^2\|=1$, so that $tx+(1-t)x^2\in \partial\Omega$, which contradicts strict convexity.

$\endgroup$
  • $\begingroup$ second line, its should be $x,y \in \Omega$, right? $\endgroup$ – JustANoob Feb 2 '18 at 15:21
  • $\begingroup$ It is the same. Indeed if $x,y\in int(\Omega)$ then $tx+(1-t)y\in int(\Omega)$ is automatically satisfied. To give you an idea about why, suppose not, so that $tx+(1-t)y\not\in int(\Omega)$. Then find $h$, $\|h\|$ small enough, so that $tx+(1-t)y+h\not\in \Omega$. But since its norm can be as small as we want, we find that $x+h,y+h\in \Omega$, and then $$ tx+(1-t)y+h = t(x+h) + (1-t)(y+h) \in \Omega, $$ which contradicts the fact that it is not there. $\endgroup$ – Tommaso Seneci Feb 2 '18 at 15:39
1
$\begingroup$

If $B=\{ f : [0,1] \mapsto \mathbb R ; \lVert f\rVert_{\infty}\leq 1 \}$ would be strictly convex, you would have for $f \neq g$ with $\Vert f \Vert_\infty = \Vert g \Vert_\infty=1$ and $0 < \alpha <1$: $$\Vert \alpha f + (1-\alpha) g \Vert_\infty <1.$$

However take $f(x) =x$ and $g(x) = x^2$. You have $\Vert f \Vert_\infty = \Vert g \Vert_\infty=1$. And for all $0 < \alpha < 1$, $h_\alpha(1)=1$ where $h_\alpha = \alpha f + (1-\alpha) g$. Hence $\Vert h_\alpha \Vert \ge 1$, in contradiction with what is required for strict convexity.

$\endgroup$
  • $\begingroup$ with $h_{a}(1)$ you mean setting $f(x)=1$ and $g(x)=1$, right? And since f and g belongs to the space, and h is not contained in the interior, the set is not strictly convex. Right? $\endgroup$ – JustANoob Feb 2 '18 at 15:14
  • $\begingroup$ Yes. That's what I mean. Please look at the definition of $h_\alpha $ in my answer. $\endgroup$ – mathcounterexamples.net Feb 2 '18 at 15:16

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.