Proof of A Weighted Young Inequality and the Splitting of a Quadratic Form

Throughout we use the notation of the statement, and we use freely that for $\alpha\le\beta$ and $0<\gamma$ one has $\gamma\alpha\le\gamma\beta$: this is immediate if $\alpha=\beta$, and otherwise is claim 10 of Elementary Order Arithmetic in an Ordered Field.

Claim 1. Let $a,b\in\mathbb{R}^{n}$ and let $s\in\mathbb{R}$. By claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and claims 1, 2, 3, 4 and 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite,

$$0\le\lVert sa-b\rVert^{2}=(sa-b)\cdot(sa-b)=s^{2}\,\lVert a\rVert^{2}-2s\,(a\cdot b)+\lVert b\rVert^{2},$$

where $s^{2}=s\cdot s$ and the nonnegativity is claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square together with claim 5 of Elementary Order Arithmetic in an Ordered Field. Hence

$$2s\,(a\cdot b)\le s^{2}\lVert a\rVert^{2}+\lVert b\rVert^{2}.$$

Now let $t\in\mathbb{R}$ be positive and take $s=t$; multiplying the resulting inequality by the positive number $t^{-1}$, which exists by claim 7 of Elementary Order Arithmetic in an Ordered Field, and using $t^{-1}t^{2}=t$ gives

$$2\,(a\cdot b)\le t\,\lVert a\rVert^{2}+t^{-1}\lVert b\rVert^{2}.$$

Claim 2. By Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (ii) the matrix $B^{2}$ lies in $\mathcal{S}(n)$. By claim 1 of The Positive Semidefinite Ordering is a Partial Order Compatible with the Linear Structure the scalar multiple $\varepsilon B^{2}$ and then the sum $B+\varepsilon B^{2}$ are again symmetric, so $B+\varepsilon B^{2}\in\mathcal{S}(n)$.

Claim 3. Let $x,z\in\mathbb{R}^{n}$ and put $h=x-z$, so that $x=z+h$. Since $B^{\top}=B$, claim 5 of Elementary Properties of the Transpose of a Real Matrix together with claim 1 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite gives $z\cdot(Bh)=h\cdot(Bz)$. Using $B(z+h)=Bz+Bh$, which is claim 3 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, and expanding with claims 2 and 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite,

$$x\cdot(Bx)=(z+h)\cdot\bigl(B(z+h)\bigr)=z\cdot(Bz)+2\,h\cdot(Bz)+h\cdot(Bh).$$

Apply claim 1 with $a=h$, $b=Bz$ and $t=\varepsilon^{-1}$, noting that $\varepsilon^{-1}$ is positive and that its multiplicative inverse is $\varepsilon$; this gives

$$2\,h\cdot(Bz)\le\varepsilon^{-1}\lVert h\rVert^{2}+\varepsilon\,\lVert Bz\rVert^{2},$$

and $\lVert Bz\rVert^{2}=z\cdot(B^{2}z)$ by Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (ii). Moreover, by claim 2 of Properties of the Norm of a Symmetric Real Matrix and claim 3 of Properties of the Absolute Value in an Ordered Field,

$$h\cdot(Bh)\le\bigl|h\cdot(Bh)\bigr|\le\lVert B\rVert\,\lVert h\rVert^{2}.$$

Combining the three displays,

$$x\cdot(Bx)\le z\cdot(Bz)+\varepsilon\,z\cdot(B^{2}z)+\varepsilon^{-1}\lVert h\rVert^{2}+\lVert B\rVert\,\lVert h\rVert^{2}.$$

Finally, by claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite,

$$z\cdot\bigl((B+\varepsilon B^{2})z\bigr)=z\cdot\bigl(Bz+\varepsilon(B^{2}z)\bigr)=z\cdot(Bz)+\varepsilon\,z\cdot(B^{2}z),$$

while distributivity in $\mathbb{R}$ gives $\varepsilon^{-1}\lVert h\rVert^{2}+\lVert B\rVert\,\lVert h\rVert^{2}=(\varepsilon^{-1}+\lVert B\rVert)\lVert h\rVert^{2}$. Substituting these two identities and $h=x-z$ into the previous display gives the asserted inequality.