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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'.

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  • $\begingroup$ Which norm on C(X)? The sup norm? $\endgroup$ – DanielWainfleet May 31 '16 at 2:49
  • $\begingroup$ You haven't told us anything about $\mu$ or $X.$ $\endgroup$ – zhw. May 31 '16 at 19:54
  • $\begingroup$ What I have stated is every thing provided in the question. I assume the norm on X is the sup norm. $\endgroup$ – user343488 May 31 '16 at 22:52
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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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    $\begingroup$ The empty set is typically written as $\emptyset$ (\emptyset) or $\varnothing$ (\varnothing), rather than $\phi$. $\endgroup$ – Rahul May 31 '16 at 3:03
  • $\begingroup$ @Rahul . 5000 books use $\phi$ for the empty set. $\endgroup$ – DanielWainfleet May 31 '16 at 3:22
  • $\begingroup$ @user254665 Name one. $\endgroup$ – user147263 May 31 '16 at 3:47
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Hint for the $C(X)$ question: Suppose $X$ is a compact metric space with more than one point. (Note that if $X$ is not compact, it's not clear that $\| \,\|_\infty$ is a norm on $C(X).$) Consider $f(x) = 1/(1+d(x,a)),g(x) = 1/(1+d(x,a))^2.$

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