# Why is the $L_p$ norm strictly convex for $1<p<\infty?$

Let $x,y \in L_p$ such that $\|x\|_p=\|y\|_p=1$ , $1< p<\infty$ and $x\neq y.$ Why is $\|x+y\|_p<2$ ?

I'm not sure how to start the proof.. I don't know how to handle integral of $(x+y)^p$ and it seems that using the binomial theorem won't be a great success.

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Start using the proof of the triangle inequality, and look at the equality case in Hölder inequality (you may have to use the strict convexity of the exponential). –  Davide Giraudo Nov 8 '11 at 10:28
Minkowski's Inequality gives $\| x+y \|_p \leq \|x\|_p + \| y \|_p = 2$ with equality if and only if $x$ and $y$ are positively linearly dependent (i.e. $x=ty$ or $y=tx$ for some $t\geq 0$).
If $x$ and $y$ were positively linearly dependent, say $x=ty$, then plugging into $\|x\|_p=\|y\|_p=1$ forces $t=1$ so $x=y$, which contradicts the assumption $x\neq y.$ Thus we do not reach the equality case, so $\| x+y \|_p <2.$