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Let $x\neq y$ when $x,y\in H$ and H is a Hilbert space which satisfy $\|x\|=\|y\|=r$. Show that $\|\frac{x+y}{2}\|<r$.

Actually in my question r=1 but as far as i could understand there is a way to prove this for any r. Is this true?

I tried to go with showing that $\|tx+(1-t)y\|<r$ where $t=\frac{1}{2}$ to no avail. Am I in a good direction? can someone point out the trick?

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    $\begingroup$ I would try using the triangle inequality, though I could be wrong. $\endgroup$ – shai horowitz May 19 '16 at 17:32
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    $\begingroup$ Hint: $||x+y||^2+||x-y||^2=2||x||^2+2||y||^2$. $\endgroup$ – David C. Ullrich May 19 '16 at 17:33
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Use paralleogram law for $\frac{x}{2}$ and $\frac{y}{2}$ to obtain $||\frac{x+y}{2}||^2 + ||\frac{x-y}{2}||^2 = \frac{1}{2}||x||^2 + \frac{1}{2}||y||^2$ and so you get $1$ in the right hand side. Since the LHS is a sum of two non-negative terms, you get the desired inequality since $x\neq y$ .

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  • $\begingroup$ Left Hand Side. $\endgroup$ – joy May 19 '16 at 17:44
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So this is my final answer for any interested party answer and thank you @joy

$\|\frac{x+y}{2}\|^2 \leq \|\frac{x+y}{2}\|^2 + \|\frac{x-y}{2}\|^2 = \frac{1}{2}\|x\|^2+\frac{1}{2}\|y\|^2$

and since $x \neq y$:

$\|\frac{x+y}{2}\|^2 < \frac{1}{2}\|x\|^2+\frac{1}{2}\|y\|^2=\frac{1}{2}r^2+\frac{1}{2}r^2=r^2$

Hence:

$\|\frac{x+y}{2}\| < r$

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    $\begingroup$ Yes... this proves for any $r$ $\endgroup$ – joy May 19 '16 at 17:55

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