Strict inequality in a Hilbert Space

Question

Let $H$ be a Hilbert Space and let $x$ and $y$ be two distinct points in $H$ such that $||x|| = ||y|| = 1$. Then show that $||tx+(1-t)y||<1$, where $0<t<1$.

It is pretty clear that any point lying on the line joining two distinct points on a sphere which is not an end-point itself will lie inside the sphere. But I have problem deducing the strict inequality here. We see that:

$$\begin{align} ||tx+(1-t)y||^2 = \langle tx+(1-t)y, tx+(1-t)y\rangle = t^2||x||^2+2t(1-t)Re\langle x,y\rangle +(1-t)^2||y||^2 \leq (t+(1-t))^2 = 1 \end{align}$$.

So, my question is how to get rid of the equality. I can not argue that $x$ and $y$ are linearly independent as these may be diametrically opposite points (on the sphere) and hence linearly dependent.

Answer 1

This is actually a problem about inner products.

The last inequality is actually Cauchy-Schwarz inequality.

$$|\langle x,y\rangle| \le ||x|| \, ||y||$$ Equality holds iff $x$ and $y$ are dependent.

However, the given conditions that $x \ne y$ and $||x||=||y||=1$ leave us two possibilities:

• $x$ and $y$ are linearly independent, so $$\Re\langle x,y\rangle \le |\langle x,y\rangle| < ||x|| \, ||y|| = 1.$$ Put this strict inequality into to your step, and you get your desired conclusion.
• $y = kx$: $||y|| = 1 \implies |k| = 1$
  • in $\Bbb{R}$, that's $x$ and $y$ are on the opposite poles of a unit sphere. Use the given condition $t \in (0,1)$ to finish this proof.
  • in $\Bbb{C}$, $\Re\langle x,y\rangle=\Re\left(\bar{k}\right) = \Re(k) \le 1$. Use $x \ne y$ to make the previous inequality strict: $$||x-y|| = ||(k-1)x|| = |k-1| \ne 0 \implies k \ne 1$$ Put $\Re\langle x,y\rangle = \Re(k) < 1$ back into your proof, and we're done.

Remarks: When you are stuck, think of which condition(s) is(are) unused.