Let $\mathcal{H}$ be a real Hilbert space and $x,y,z \in \mathcal{H}$. If $x$ and $y$ are a convex combination of $z$, that is
$$ z := tx + (1-t)y , \quad t \in (0,1) , $$
then we have the following inequality
$$ \left\lVert z \right\rVert ^{2} \leq t \left\lVert x \right\rVert ^{2} + (1-t)\left\lVert y \right\rVert ^{2} \tag{1}\label{1} . $$
Notice that \eqref{1} is better than the Cauchy – Schwarz inequality in the sense that both coefficients are strictly less than $1$.
The question is when we only have
$$ \left\langle z , v \right\rangle = \left\langle tx + (1-t)y , v \right\rangle , \quad t \in (0,1) , \tag{2}\label{2} $$
and $v \in \mathcal{H}$ is arbitrary given, can we still derive the inequality \eqref{1}.
A naive approach gives
$$ \left\lVert z \right\rVert ^{2} – \left\lVert z-v \right\rVert ^{2} = t \left\lVert x \right\rVert ^{2} + (1-t) \left\lVert y \right\rVert ^{2} – t (1-t) \left\lVert x-y \right\rVert ^{2} – \left\lVert tx + (1-t)y-v \right\rVert ^{2} $$
If we can not get \eqref{1}, is there any alternative one in which I can avoid or at least weaken the term $- \left\lVert z-v \right\rVert ^{2}$, that is, to have $-c \left\lVert z-v \right\rVert ^{2}$ for some $c \in (0,1)$?
Best Answer
It fails in $\mathcal{H}=\mathbb{R}^3$ with the usual inner product. Let's consider the following particular case: $y=(0,0,0)$, $x=(2,0,0)$, $z=(\lambda,0,0)$ with $\lambda^2>2$, $v=(0,1,0)$ and $t=1/2$. Then $\langle z,v \rangle =0 =\langle tx+(1-t)y,v\rangle $ but $||z||^2=\lambda^2>2=t||x||^2$.