Hilbert space Inequality

Question

Let $K ⊂H$ be a nonempty closed convex set, $H$ Hilbert. Let $f ∈H$ and let $u=P_Kf$. Prove that $$||v − u||^2 ≤ ||v − f ||^2 − ||u − f ||^2, ∀v ∈ K$$

I've tried to use the parallelogram identity and the relation $||u-v||^2=||u||^2+||v||^2-2<u,v>$ without any success.

Do you have any hint?

This is what I have done, but it does not work!

$\forall v,u\in H$ it holds: $||v+u||^2=||u||^2+||v||^2+2<u,v>$ (*),

then:

$||v-u||^2=||v-f+f-u||^2=$

$=||(v-f)+(f-u)||^2=^{(*)}$

$=||u-f||^2+||v-f||^2+2<v-f,f-u>$

$=||u-f||^2+||v-f||^2-2<f-v,f-u>$

but now?

Answer 1

$\forall v,u\in H$ it holds: $||v+u||^2=||u||^2+||v||^2+2<u,v>$ (*),

then:

$||v-u||^2=||v-f+f-u||^2=$

$=||(v-f)+(f-u)||^2=^{(*)}$

$=||u-f||^2+||v-f||^2+2<v-f,f-u>=$(1)

Now we have: $<v-f,f-u>=<v-u+u-f,f-u>=<v-u,f-u>-||u-f||^2$, therefore:

(1)$=||u-f||^2+||v-f||^2+2(<v-u,f-u>-||u-f||^2)=$

$=||v-f||^2-||u-f||^2+2<v-u,f-u>.$

To summarise:

$$||v-u||^2=||v-f||^2-||u-f||^2+2<v-u,f-u>$$

Since $u=P_Kf$ it holds: $⟨f−u,v−u⟩≤0, \forall v\in H$, it is now straightforward to see that:

$||v-u||^2\leq ||u-f||^2+||v-f||^2.$

Answer 2

Let me also give a hint on how to start:

$$\|v -P_K f \|^2 = \langle v -f +f -P_K f, v- P_K f\rangle = \langle v -f, v - P_K f\rangle + \langle f -P_K f, v - P_K f\rangle$$

With my hint, i.e. $\langle f -P_K f, v -P_K f\rangle\leq 0$ (at least in a real Hilbert-space, otherwise this only holds for the real part of the expression) applied to the second term of the rhs and with your parallelogramm identity applied to the first term of the rhs, we obtain

$$\|v -P_K f \|^2 \leq \frac{1}{2}\left(\|v-f\|^2 + \|v -P_K f\|^2 -\|f-P_Kf\|^2 \right)$$ Multiply by $2$ and subtract the term $\|v-P_K f\|^2$ to complete the proof.