Prove that a set is not strictly convex

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

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

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.
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.