Ficken Characterization : Given X is normed space. For any vectors x,y in X and if $\Vert x \Vert = \Vert y\Vert$, then $\Vert ax+by\Vert = \Vert bx + ay \Vert$ , for any scalars a & b. The conclusion: X is inner product space. Ficken modify $\Vert x+y \Vert = \Vert x-y+2y \Vert$ and $\Vert y-x+2x \Vert$, $\Vert x+y-2y\Vert$, $\Vert x+y-2x \Vert$.
On page 364, Ficken obtains 4 equation:
- $\Vert y \Vert \Vert x+y\Vert\Vert x-y \Vert = \big\Vert 2 \Vert y \Vert^2x+(\Vert x-y \Vert^2 - 2 \Vert y \Vert^2)y \big \Vert$
- $\Vert y\Vert \Vert x+y\Vert \Vert x-y\Vert = \big \Vert 2\Vert y\Vert ^2x + (\Vert x-y\Vert^2 - 2\Vert x \Vert^2)y \big \Vert$
- $\Vert y\Vert \Vert x+y\Vert \Vert x-y\Vert = \big \Vert 2\Vert y\Vert ^2x + (2\Vert y \Vert^2 -\Vert x+y\Vert^2)y \big \Vert$
- $\Vert y\Vert \Vert x+y\Vert \Vert x-y\Vert = \big \Vert 2\Vert y\Vert ^2x + (2\Vert x \Vert^2 -\Vert x+y\Vert^2)y \big \Vert$
for all x,y.
Especially on page 364 second and fourth equation, I don't know how to get it. Can somebody help me? Thanks in advanced.
PS: You can check a full paper version in here https://sci-hub.tw/10.2307/1969273.
On the first equation, we can modify $\Vert x+y \Vert = \Vert x-y + 2y \Vert = \big\Vert (x-y) + 2\frac{\Vert y \Vert \Vert x-y\Vert}{\Vert y \Vert \Vert x-y \Vert} y \big\Vert$. Because $\Vert x-y \Vert = \big\Vert \frac{\Vert x-y \Vert}{\Vert y \Vert} y \big\Vert$, then we can do, $\big \Vert \frac{2\Vert y \Vert}{\Vert x-y \Vert}(x-y) + \frac{\Vert x - y \Vert}{\Vert y \Vert}y \big \Vert$. By algebraic manipulation we get $\Vert x + y \Vert = \frac{\Vert 2 \Vert y \Vert^2 x + (\Vert x-y \Vert^2 - 2 \Vert y \Vert^2) y \Vert}{\Vert y \Vert \Vert x-y \Vert}$.
On the third equation can be solved with same trick. But I still confuse how to get second and fourth equation. I want to show second equation and fourth equation, for all x,y.
I have already found the way.
On the second equation we can write: $\Vert x+y \Vert = \Vert (y-x) + 2x \Vert = \Vert (y-x) + 2\frac{\Vert y-x\Vert \Vert x \Vert}{\Vert y-x\Vert \Vert x \Vert}x \Vert$. With the same trick like the first equation, we get $\Vert x \Vert \Vert x+y \Vert \Vert x-y \Vert = \Big\Vert 2 \Vert x \Vert^2y + (\Vert x-y\Vert^2 - 2\Vert x \Vert^2)x \Big\Vert.$ Multiply with $\frac{\Vert y \Vert}{\Vert x \Vert}$, and we obtain $\Vert y \Vert \Vert x+y \Vert \Vert x-y \Vert = \Big\Vert 2 \Vert y \Vert \Vert x \Vert y + (\Vert x-y\Vert^2 - 2\Vert x \Vert^2)x \frac{\Vert y \Vert}{\Vert x \Vert}\Big\Vert.$
Because $\Vert y \Vert = x \frac{\Vert y \Vert}{\Vert x \Vert}$, we can switch the constant. Finally $ \Vert y \Vert\Vert x+y \Vert \Vert x-y \Vert = \Big\Vert 2 \Vert y \Vert\Vert x \Vert \frac{\Vert y \Vert}{\Vert x \Vert}x+ (\Vert x-y \Vert^2 - 2 \Vert x \Vert^2) y \Big\Vert.$ $ \Vert y \Vert\Vert x+y \Vert \Vert x-y \Vert = \Big\Vert 2 \Vert y \Vert^2 x + (\Vert x-y \Vert^2 - 2 \Vert x \Vert^2) y \Big\Vert.$
And the second equation is proved. With the same trick, we can obtain the fourth equation. Thank you.