1. open ball

Let $\displaystyle |x|=|y|=r$ with $\displaystyle x \neq y$. If $\displaystyle 0<t<1$ prove that $\displaystyle |(1-t)x+ty|<r$.

I've proven that $\displaystyle |(1-t)x+ty| \leq r$. Now how do I show that this inquality is strict?

2. Originally Posted by jefferson_lc
Let $\displaystyle |x|=|y|=r$ with $\displaystyle x \neq y$. If $\displaystyle 0<t<1$ prove that $\displaystyle |(1-t)x+ty|<r$.
I've proven that $\displaystyle |(1-t)x+ty| \leq r$. Now how do I show that this inquality is strict?
In the post that I deleted I was going by the title of the post.
Are you trying to prove that an open ball is convex?
Did you post exactly what you meant to post?

3. Originally Posted by Plato
In the post that I deleted I was going by the title of the post.
Are you trying to prove that an open ball is convex?
Did you post exactly what you meant to post?
I've posted exatcly what I meant to post. Nevermind the title. I couldn't come up with a decent one so I wrote "open ball". I'm sorry for the misleading title.

4. Originally Posted by jefferson_lc
Let $\displaystyle |x|=|y|=r$ with $\displaystyle x \neq y$. If $\displaystyle 0<t<1$ prove that $\displaystyle |(1-t)x+ty|<r$.

I've proven that $\displaystyle |(1-t)x+ty| \leq r$. Now how do I show that this inquality is strict?
You need to give us a bit more context here. What are x and y supposed to be – real numbers, complex numbers, vectors in a euclidean space, or what?

In fact, in every case the method is the same: form the square $\displaystyle |(1-t)x+ty|^2$. Expand that, either as a product or in terms of inner products, depending on what sort of space you are working in, and see when that can be equal to $\displaystyle r^2$. You should end up with an equation like $\displaystyle \langle x,y\rangle = r^2$ (again, depending on the space). You then need to show why that can only happen when $\displaystyle x = y$.