When reading "Chebyshev centers and uniform convexity" by Dan Amir I encountered the following results which is apparently "known and easy to prove". I'm sure it is, but I can't find a proof and am failing to prove it myself.

The result (slightly simplified) is

If $X$ is a uniformly convex space (i.e. if $||x|| = ||y|| = 1$ with $||x - y|| \geq \epsilon$ then there exists $\delta(\epsilon) > 0$ such that $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$) then for any $x, y$ with $||x|| \leq 1$ and $||y|| \leq 1$, and $||x - y|| \geq \epsilon$, $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$.

Part of the problem is that I think this isn't true without making some additional restrictions to reduce the value of $\delta(\epsilon)$. e.g. by considering $||x|| = 1$ and $y = (1 - \epsilon) x$ you can see that this requires that $\delta(\epsilon) \leq \frac{1}{2} \epsilon$. So I think the true result is probably just that you can choose $\delta$ so that this is true.

I'm sure this should be easy and I'm just missing an obvious trick, but oh well.

When reading "Chebyshev centers and uniform convexity" by Dan Amir I encountered the following result which is apparently "known and easy to prove". I'm sure it is, but I can't find a proof and am failing to prove it myself.

The result (slightly simplified) is

If $X$ is a uniformly convex space (i.e. if $||x|| = ||y|| = 1$ with $||x - y|| \geq \epsilon$ then there exists $\delta(\epsilon) > 0$ such that $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$) then for any $x, y$ with $||x|| \leq 1$ and $||y|| \leq 1$, and $||x - y|| \geq \epsilon$, $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$.

Part of the problem is that I think this isn't true without making some additional restrictions to reduce the value of $\delta(\epsilon)$. e.g. by considering $||x|| = 1$ and $y = (1 - \epsilon) x$ you can see that this requires that $\delta(\epsilon) \leq \frac{1}{2} \epsilon$. So I think the true result is probably just that you can choose $\delta$ so that this is true.

I'm sure this should be easy and I'm just missing an obvious trick, but oh well.

When reading "Chebyshev centers and uniform convexity" by Dan Amir I encountered the following results which is apparently "known and easy to prove". I'm sure it is, but I can't find a proof and am failing to prove it myself.

The result (slightly simplified) is

If $X$ is a uniformly convex space (i.e. if $||x|| = ||y|| = 1$ with $||x - y|| \geq \epsilon$ then there exists $\delta(\epsilon) > 0$ such that $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$) then for any $x, y$ with $||x|| \leq 1$ and $||y|| \leq 1$, $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$.

Part of the problem is that I think this isn't true without making some additional restrictions to reduce the value of $\delta(\epsilon)$. e.g. by considering $||x|| = 1$ and $y = (1 - \epsilon) x$ you can see that this requires that $\delta(\epsilon) \leq \frac{1}{2} \epsilon$. So I think the true result is probably just that you can choose $\delta$ so that this is true.

I'm sure this should be easy and I'm just missing an obvious trick, but oh well.

When reading "Chebyshev centers and uniform convexity" by Dan Amir I encountered the following results which is apparently "known and easy to prove". I'm sure it is, but I can't find a proof and am failing to prove it myself.

The result (slightly simplified) is

If $X$ is a uniformly convex space (i.e. if $||x|| = ||y|| = 1$ with $||x - y|| \geq \epsilon$ then there exists $\delta(\epsilon) > 0$ such that $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$) then for any $x, y$ with $||x|| \leq 1$ and $||y|| \leq 1$, and $||x - y|| \geq \epsilon$, $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$.

Part of the problem is that I think this isn't true without making some additional restrictions to reduce the value of $\delta(\epsilon)$. e.g. by considering $||x|| = 1$ and $y = (1 - \epsilon) x$ you can see that this requires that $\delta(\epsilon) \leq \frac{1}{2} \epsilon$. So I think the true result is probably just that you can choose $\delta$ so that this is true.

I'm sure this should be easy and I'm just missing an obvious trick, but oh well.