# Matrix integral inequality with Frobenius norm

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?

• Use the definition of the norm (squared) and deal with the terms individually.
– KBS
Apr 10 at 14:30

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.

• I didn't expect this question to be so simple. . . Apr 11 at 0:41

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