Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question seems basic but I could not find an answer. I have seen the inequality $$\left\|\int_a^b x(t) dt \right\| \leq \int_a^b \left\| x(t) \right\| dt $$ where $x(t) \in \mathbb{R}^n$ is a vector function and $\|\cdot\|$ is a vector norm, and $a < b$.

I wonder if this also holds for matrices with induced norm, that is $$\left\|\int_a^b X(t) dt \right\| \leq \int_a^b \left\| X(t) \right\| dt $$ where $X(t)$ is a matrix function and $\|\cdot\|$ is an induced matrix norm, and $a < b$. If it is true, is there any reliable citation source?

share|improve this question
    
By induced norm, you mean a norm of the form $\sup_{v\neq 0}\frac{||Av||}{||v||}$? –  Davide Giraudo Feb 29 '12 at 20:41
    
The space $M_n(\mathbb{R})$ is just a finite dimensional vector space like any other, so you can directly apply the the triangle inequality to any matrix-valued function, with any norm, induced or not (actually you could even have non square matrices). Induced norm are useful when dealing with matrix product but there's none here. –  Joel Cohen Mar 1 '12 at 0:51
add comment

2 Answers 2

up vote 3 down vote accepted

If Riemann-integrals are good enough for you, then these inequalities are just the disguised triangle inequality (here with sloppy notation): $$ \left\|\int_a^b X(t) dt \right\| = \left\| \lim_{\mathcal Z} \sum_{\mathcal Z}X(\xi) \right\| \leq \lim_{\mathcal Z} \sum_{\mathcal Z} \left\| X(\xi) \right\| \leq \int_a^b \left\| X(t) \right\| dt$$

share|improve this answer
    
So is this inequality true for all matrix norms, not just induced norms? –  Truong Feb 29 '12 at 21:04
    
I think so, look at math.ucsd.edu/~bdriver/231-02-03/Lecture_Notes/chap4.pdf –  Blah Feb 29 '12 at 22:05
add comment

If $||\cdot||$ is given by $||A||=\sup_{v\neq 0}\frac{N(Av)}{N(v)}$ then we have for a fixed $v$ $$N\left(\int_a^bX(t)dtv\right)=N\left(\int_a^bX(t)vdt\right)\leq \int_a^bN\left(X(t)v\right)dt\leq \int_a^b||X(t)||N(v)dt$$ so $\|\int_a^b X(t)dt\|\leq \int_a^b||X(t)||dt$.

In fact more generally, if $f\colon [a,b]\to X$ where $(X,||\dot||$ is a Banach space then $\|\int_a^b f(t)dt\|\leq \int_a^b||f(t)||dt$. It can be showed using a corollary of Hahn-Banach theorem: \begin{align*}\|\int_a^b f(t)dt\|&=\sup_{\varphi\in X',||\varphi||=1}\left|\varphi\left(\int_a^b f(t)dt\right)\right|\\ &=\sup_{\varphi\in X',||\varphi||=1}\left|\int_a^b \varphi\left(f(t)\right)dt\right|\\ &\leq \sup_{\varphi\in X',||\varphi||=1}\int_a^b \left|\varphi\left(f(t)\right)\right|dt\\ &\leq \sup_{\varphi\in X',||\varphi||=1}\int_a^b \|f(t)\|dt \\ &=\int_a^b \|f(t)\|dt. \end{align*}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.