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
 A: 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.
A: 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.
