Matrix integral inequality with Frobenius norm

Question

Let $||\cdot||$ be the matrix Frobenius norm and $A(t)$ is a matrix function. Is it always true that $$\left|\left|\int_0^1 A(t) dt \right|\right| \leq \int_0^1 \left|\left| A(t)\right|\right| dt \ ?$$

If $||\cdot||$ is an induced matrix norm, this has been proved here, but it can't be used for this problem. Can anyone help me?

Answer 1

We have that

$$ \left|\left|\int_0^1A(t)dt\right|\right|_F^2=\sum_{ij}\left|\int_0^1a_{ij}(t)dt\right|^2\le \sum_{ij}\int_0^1\left|a_{ij}(t)\right|^2dt=\int_0^1\left|\left|A(t)\right|\right|_F^2dt $$ where the inequality is an application of the Cauchy–Schwarz inequality.

Answer 2

Let $v=\operatorname{vec}(A)$. You are asking whether

$$\left\|\int_0^1 A(t) dt \right\| =\sqrt{\sum_i\left(\int_0^1 v_i(t) dt\right)^2} \le\int_0^1\sqrt{\sum_iv_i(t)^2}\,dt =\int_0^1 \left\|A(t)\right\| dt.$$

Assuming that $A(t)$ is Riemann integrable, it suffices to prove the following inequality between Riemann sums:

$$\sqrt{\sum_i\left(\frac{1}{n}\sum_j v_i(\tfrac{j}{n})\right)^2} \le\frac{1}{n}\sum_j\sqrt{\sum_iv_i(\tfrac{j}{n})^2}.$$

To prove it, cancel out the factor $\frac{1}{n}$ on both sides first. Let $V$ be the rectangular matrix such that $v_{ij}=v_i(\frac{j}{n})$. Then

\begin{align*} \sum_i\left(\sum_j v_{ij}\right)^2 &=\sum_i\sum_jv_{ij}^2+2\sum_i\sum_{j<k}v_{ij}v_{ik}\\ &=\sum_i\sum_jv_{ij}^2+2\sum_{j<k}\sum_iv_{ij}v_{ik}\\ &=\sum_i\sum_jv_{ij}^2+2\sum_{j<k}\langle v_{\ast j},\,v_{\ast k}\rangle\\ &\le\sum_j\sum_iv_{ij}^2+2\sum_{j<k}\|v_{\ast j}\|\|v_{\ast k}\|\\ &=\sum_j\sum_iv_{ij}^2+2\sum_{j<k}\sqrt{\left(\sum_iv_{ij}^2\right)\left(\sum_iv_{ik}^2\right)}\\ &=\left(\sum_j\sqrt{\sum_iv_{ij}^2}\right)^2.\quad\text{QED} \end{align*}