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 Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and claims 1, 2, 3, 4 and 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$,

where $s^{2}=s\cdot s$ and the nonnegativity is claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ together with claim 5 of Elementary Order Arithmetic in an Ordered Field. Hence

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

Claim 2. By Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix §square 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 Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ 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 Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$,

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

and $\lVert Bz\rVert^{2}=z\cdot(B^{2}z)$ by Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix §square. 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,

Combining the three displays,

Finally, by claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and claim 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$,

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.