Proof of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals

1. $|x|^{2}=\sum_l(x^{l})^{2}$ dominates each $(x^{i})^{2}$, giving $|x^{i}|\le|x|$; and $\bigl(\sum_l|x^{l}|\bigr)^{2}=\sum_l(x^{l})^{2}+\sum_{l\ne l'}|x^{l}||x^{l'}|\ge|x|^{2}$, giving $|x|\le\sum_l|x^{l}|$, using that the nonnegative square root is monotone (if $0\le u\le v$ then $\sqrt u\le\sqrt v$, since otherwise squaring the reverse strict inequality contradicts $u\le v$); the final bound replaces each summand by the maximum. The matrix statements are the vector statements for the tuple of $pq$ entries, together with $\sum_{i,j}X_{ij}^{2}\le pq\,|X|_{e}^{2}$ and monotonicity of the square root ($\sqrt{pq}\le pq$ since $pq\ge1$).

2. $|(UV)_{il}|=\bigl|\sum_{j=1}^{q}U_{ij}V_{jl}\bigr|\le\sum_{j=1}^{q}|U_{ij}|\,|V_{jl}|\le q\,|U|_{e}\,|V|_{e}$ by the triangle inequality for real numbers.

3. Entrywise, $((UV)^{\top})_{ij}=(UV)_{ji}=\sum_lU_{jl}V_{li}=\sum_l(V^{\top})_{il}(U^{\top})_{lj}=(V^{\top}U^{\top})_{ij}$ with the transpose. Similarly, both $y\cdot(Mz)$ and $(M^{\top}y)\cdot z$ equal the double sum $\sum_{i,l}y^{i}M_{il}z^{l}$, by unfolding the matrix-vector product and dot product and exchanging the two finite summations.

4. For $a\le s\le t\le b$: if $s=t$ both sides of the first identity agree by the degenerate-interval convention of Mean-Square Riemann Integral of a Family of Random Variables; if $a=s<t$ it is trivial (again using the convention for $\int_a^a$); and if $a<s<t$ it is Additivity of the Riemann Integral on Adjacent Intervals applied on $[a,t]$ with intermediate point $s$, rearranged. For the bound: $|\varphi|$ is continuous (for any $u,u'$, $\bigl||\varphi(u)|-|\varphi(u')|\bigr|\le|\varphi(u)-\varphi(u')|$ by the triangle inequality for reals), so $M:=\max_{[a,b]}|\varphi|$ exists by Extreme Value Theorem on a Compact Subset of a Metric Space, the interval $[a,b]$ being nonempty, since $a<b$, and a compact subset of the real line by Closed Interval $[a,b]$ is Compact in $\mathbb{R}$; by monotonicity and linearity of the Riemann integral on continuous integrands (via claim 3 of the integral toolkit on a compact interval, which identifies that integral with the Lebesgue integral, together with Linearity and Monotonicity of the Lebesgue Integral), $\pm\int_s^t\varphi\le\int_s^t|\varphi|\le M(t-s)$ for $s<t$, and the degenerate case is $0$. Continuity of the indefinite integral at every point of $[a,b]$ is then immediate: $\bigl|\int_a^t\varphi-\int_a^s\varphi\bigr|\le M\,|t-s|$.

5. Continuity of $|w^{i}(\cdot)|$ is as in claim 4; continuity of $|w(\cdot)|$ follows since, by claim 1 and the triangle inequality in $\mathbb{R}^{p}$ (Euclidean Distance is a Metric on $\mathbb{R}^n$), $\bigl||w(r)|-|w(r')|\bigr|\le|w(r)-w(r')|\le\sum_i|w^{i}(r)-w^{i}(r')|$. The first inequality of the display is claim 1 applied to the vector of componentwise integrals together with $\bigl|\int_a^tw^{i}\bigr|\le\int_a^t|w^{i}|$ (as in claim 4); the second uses $|w^{i}(r)|\le|w(r)|$ (claim 1) and monotonicity of the integral, summed over $i$.

6. Unfolding as in claim 3, the left side equals $\sum_{i,l}y^{i}z^{l}\int_a^tU_{il}(r)\,dr$, and by linearity of the Riemann integral on continuous integrands (claim 3 of the integral toolkit on a compact interval, which identifies that integral with the Lebesgue integral, together with Linearity and Monotonicity of the Lebesgue Integral) this equals $\int_a^t\sum_{i,l}y^{i}U_{il}(r)z^{l}\,dr=\int_a^ty\cdot(U(r)z)\,dr$; degenerate $t=a$ by the convention.

7. Continuity of $\tau\mapsto u(a+\tau)$ on $[0,b-a]$ is immediate from continuity of $u$ (the shift is distance-preserving). For $t=a$ both sides are $0$ by the convention. For $t>a$, write $I=\int_a^tu(r)\,dr$ and let $\varepsilon>0$; take $\delta>0$ from Riemann Integrability on a Closed Interval for $u$ on $[a,t]$. Every tagged partition $(\tau_0,\dots,\tau_n)$ of $[0,t-a]$ with tags $\sigma_i$ and mesh less than $\delta$ shifts to the tagged partition $(a+\tau_0,\dots,a+\tau_n)$ of $[a,t]$ with tags $a+\sigma_i$ and the same mesh, and the two Riemann sums are equal term by term: $\sum_iu(a+\sigma_i)(\tau_i-\tau_{i-1})=\sum_iu(a+\sigma_i)\bigl((a+\tau_i)-(a+\tau_{i-1})\bigr)$. Hence every Riemann sum of $\tau\mapsto u(a+\tau)$ on a tagged partition of $[0,t-a]$ of mesh less than $\delta$ lies within $\varepsilon$ of $I$. Since $\tau\mapsto u(a+\tau)$ is continuous, it is Riemann integrable by claim 3 of the integral toolkit on a compact interval, and its integral $J$ is, by Riemann Integrability on a Closed Interval, also approximated within any $\varepsilon'>0$ by all sums of sufficiently small mesh; taking a common sufficiently fine tagged partition gives $|I-J|\le\varepsilon+\varepsilon'$ for all $\varepsilon,\varepsilon'>0$, so $J=I$. $\blacksquare$