When $1 , $||f||_p=1$,$||g||_p=1$,$f\ne g$,then $\frac{1}{2} ||f+g||_p<1$. I use parallelogram law $||f+g||^2+||f-g||^2=||f||^2+||g||^2=4\\$ Since $f\ne g$, $||f-g||^2>0$ then$||f+g||_p<1$ But my proof does not use $1 , and this obviously fail when p is 1 or $\infty$. I just want to know where my proof is wrong. I guess is this because the step Since $f\ne g$,$||f-g||^2>0$ ?I just use the definition 5.2 in rudin's book (c)$||x||=0 $ implies $x=0$.
Prove unit ball of $L^p(\mu)$ is strictly convex, when 1
2
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real-analysis
norm
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2The parallelogram law is only valid on inner product spaces. $L^p$ when $p\neq 2$ is not an inner product space, so you'll have to use another method. – 2012-11-12
1 Answers
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Note that norm satisfies the triangle inequality by the definition of norm, $|f|_p+|g|_p\geq |f+g|_p$
So if $|g|_p$, $|f|_p \leq 1$, then for $0< t< 1$, $ 1=(1-t) + t \geq |(1-t)f|_p+|tg|_p \geq |(1-t)f+tg|_p$