Inequality for convex combination inner product

Question

Let $\mathcal{H}$ be a real Hilbert space and $a,b,c,d \in \mathcal{H}, t \in (0,1)$ be such that $$ \left\langle a - t \cdot c , a - b \right\rangle = (1-t) \left\langle b - t \cdot d , a - b \right\rangle + t \left\langle - t \cdot d , a - b \right\rangle \neq 0 \tag{1}\label{1} . $$ Notice that if we set $$ v := a-b, x:= b - t \cdot d , y:= - t \cdot d, z := a - t \cdot c , $$ then we have $$ \left\langle z , v \right\rangle = (1-t) \left\langle x , v \right\rangle + t \left\langle y , v \right\rangle = \left\langle (1-t)x + ty , v \right\rangle . $$ This reminds us about the convex combination $$ z = (1-t)x + ty $$ in which we can get the following inequality $$ \left\lVert z \right\rVert ^{2} \leq (1-t) \left\lVert x \right\rVert ^{2} + t \left\lVert y \right\rVert ^{2} \tag{2}\label{2} . $$ The question is can we derive some similar inequality like \eqref{2} for \eqref{1}. Otherwise, as a naive approach gives \begin{align*} \left\lVert a - t \cdot c \right\rVert ^{2} - \left\lVert b - t \cdot c \right\rVert ^{2} & = (1-t) \left\lVert b - t \cdot d \right\rVert ^{2} + t \left\lVert - t \cdot d \right\rVert ^{2} - t (1-t) \left\lVert b \right\rVert ^{2} \\ & \qquad - \left\lVert (2-t)b - t \cdot d - a \right\rVert ^{2} . \end{align*} I would be happy to weaken the term $- \left\lVert b - t \cdot c \right\rVert ^{2}$, that is to have $- \delta \left\lVert b - t \cdot c \right\rVert ^{2}$ for some $\delta \in (0,1)$.

Answer 1

Unfortunately, in this imperfect world we not always can be happy.

Simplifying each side in the last equality in your question we obtain

$$\langle a,a\rangle-2t\langle a,c\rangle+t^2\langle c,c\rangle-[\langle b,b\rangle-2t\langle b,c\rangle+t^2\langle c,c\rangle]\le$$ $$\langle a,a\rangle+2(2-t) \langle a,b\rangle-2t \langle a,d\rangle+(2t-3) \langle b,b\rangle+2t\langle b,d\rangle.$$

As I understood you, it should be an inequality with some multiplier $\delta$ before the square brackets, which is as small as possible. Then it is simplified to

$$0\le (4-2t) \langle a,b\rangle+2t\langle a,c\rangle-2t \langle a,d\rangle+(2t-3+\delta) \langle b,b\rangle-2t\delta\langle b,c\rangle+2t\langle b,d\rangle+t^2(\delta-1)\langle c,c\rangle\tag{3}\label{3}$$

On the other hand, (1) imply

$$ \langle a,a\rangle+(t-2)\langle a,b\rangle- t\langle a,c\rangle+t \langle a,d\rangle+(1-t) \langle b,b\rangle+t\langle b,c\rangle-t\langle b,d\rangle=0. \tag{4}\label{4}$$

Myltiplying (\ref{4}) by two and adding to (\ref{3}), we obtain

$$0\le 2\langle a,a\rangle+(\delta-1) \langle b,b\rangle+2t(1-\delta)\langle b,c\rangle+t^2(\delta-1)\langle c,c\rangle=$$ $$2\|a,a\|+(\delta-1)\|b-tc\|.\tag{5}\label{5}$$

Now for any $\delta<1$ we can easily find $a$, $b$, $c$, $d$, and $t$ satisfying (1), but violating (\ref{5}). For the simplicity put $a=d=0$, $b=-c\ne 0$ and $t=1/2$.