# Strictly convex Inequality in $l^p$

Let $1<p<\infty$. Let $x,y\in l^p$ such that $||x||_p=1$, $||y||_p=1$ and $x\neq y$. Would you help me to show that for any $0<t<1$, $||tx+(1-t)y||_p<1$.

My answer : By using Minkowski inequality, we get $||tx+(1-t)y||_p\leq t||x||_p+(1-t)||y||_p=t+(1-t)=1$. But I don't get the strict inequality.

But, For $p=2$: \begin{eqnarray} ||tx+(1-t)y||_2^2&=&t^2||x||_2^2+(1-t)^2||y||_2^2+2t(1-t)\Re(<x,y>)\\&=&1+2t(1-t)(\Re(<x,y>)-1)) \end{eqnarray}

Since $x\neq y$ and $||x||_2=||y||_2$, we conclude that $x\neq ky$ for every scalar hence we get $\Re(<x,y>)\leq|<x,y>|<||x||_2||y||_2=1$. So, $||tx+(1-t)y||_2<1$.

Thanks everyone.

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## 1 Answer

For $1 < p < \infty$, Minkowski's inequality is an equality if and only if one of the vectors is a multiple of the other by a nonnegative scalar.

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If the vectors is multiple of the other by scalar implies the equality is trivial. How about the converse? – beginner Nov 30 '12 at 6:43
You can go back over the proof of Minkowski's inequality, and Hölder's which is usually used in that proof. At some point you may use something like the fact that the minimum of the function $f(x) = x^p/p - x$ for $x \ge 0$ is at $x=1$. Now note that this is a unique minimum ($f$ is strictly convex)... – Robert Israel Nov 30 '12 at 8:20