# Prove that a set $\{ f:\lVert f\rVert_{\infty}\leq 1 \}$ is not strictly convex

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

• What is $\|f\|_{\infty}$? What is $\|g\|_{\infty}$? What is $\|\lambda f + (1 - \lambda) g\|_{\infty}$ if $0 < \lambda < 1$? – Hans Engler Feb 2 '18 at 14:47
• Check that the line between $x$ and $x^2$ lies on the boundary of your convex set. – Jeff Feb 2 '18 at 14:47
• 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. – Hyperplane Feb 2 '18 at 14:52
• $\lambda$= 1/2 does the job. Intuitively it should be as you say, i just don't think i can answer in that way. – JustANoob Feb 2 '18 at 15:01

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.

• second line, its should be $x,y \in \Omega$, right? – JustANoob Feb 2 '18 at 15:21
• 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. – Tommaso Seneci Feb 2 '18 at 15:39

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.

• 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? – JustANoob Feb 2 '18 at 15:14
• Yes. That's what I mean. Please look at the definition of $h_\alpha$ in my answer. – mathcounterexamples.net Feb 2 '18 at 15:16