[Math] Unit ball of $L^1$, $L^\infty$ and $C(X)$ is not strictly convex

banach-spacesfunctional-analysisnormed-spacesreal-analysis

I need to show that the unit balls of $L^1(\mu)$, $L^\infty(\mu)$ and $C(X)$ are not strictly convex.

I have already shown that if $1<p<\infty$ then the unit ball of $L^p(\mu)$ is strictly convex; i.e., if
$||f||_p=||g||_p=1$, $f\ne g$, $h=\frac{1}{2}(f+g)$
then $||h||_p<1$.

I have already researched similar questions but none of them explain why these three spaces are not strictly convex. Also, this is Problem 3 in Chapter 5 of Rudin's 'Real and Complex Analysis'.

Best Answer

Suppose $0<\mu (A)<\infty$ and $0<\mu (B)<\infty,$ and $A\cap B=\phi.$

(1).Let $f(x)=1/\mu (A)$ when $x\in A$ and $f(x)=0$ when $x\not \in A.$ Let $g(x)=1/\mu (B)$ when $x \in B$ and $g(x)=0$ when $x\not \in B.$ Then $f, g$ are linearly independent, and $\|(f+g)/2\|_1=\|f\|_1=\|g\|_1=1.$

(2). Let $f(x)=1$ when $x\in A$ and $f(x)=0$ when $x \not \in A.$ Let $g(x)=1$ when $x\in A\cup B$ and $g(x)=0$ when $x \not \in A\cup B.$ Then $f, g$ are linearly independent and $\|(f+g)/2\|_{\infty}=\|f\|_{\infty}=\|g\|_{\infty}=1.$

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