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Let $1. 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, $||tx+(1-t)y||_p<1$.

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()\\&=&1+2t(1-t)(\Re()-1)) \end{eqnarray}

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

Thanks everyone.

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