[Math] Inequality involving norm of matrix integral

inequalityintegrationmatricesnormed-spaces

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 $a < b$.

Best Answer

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

Related Solutions

[Math] Counterexamples to the Matrix norm AM-GM inequality

Consider $$A = \begin{pmatrix} 46 & 7 & 8 \\ 7 & 36 & 16 \\ 8 & 16 & 30\end{pmatrix}\qquad \text{ and }\qquad B=\begin{pmatrix} 34 & 7 & 12 \\ 7 & 48 & 10 \\ 12 & 10 & 44\end{pmatrix}.$$ Then $$\frac{1}{2}\|A+B\|_1 =\frac{1}{2}\|A+B\|_{\infty}=62$$ however $$ \|A^{1/2}B^{1/2}\|_1\approx 62.1527 \qquad \text{ and }\qquad \|A^{1/2}B^{1/2}\|_{\infty}\approx 62.1037.$$

Here is the Matlab code to generate such counter examples:

n = 3;  % choose the dimension of the matrices
stop = 0;  
while stop == 0  
  % Generates a pair of symmetric positive definite matrices
  A = floor(10*rand(n));  
  A=A+A';  
  A = A+10*n*eye(n);  
  B = floor(10*rand(n));  
  B = B+B';  
  B = B+10*n*eye(n);  
  % Computes their square root
  AS = A^(1/2);  
  BS=B^(1/2);
  % Check if the AM-GM is not satisfied
  if norm(AS*BS,1)>norm(A+B,1)/2 && norm(AS*BS,inf)>norm(A+B,inf)/2  
    stop =1;  
  end  
end

[Math] Matrix norm inequality proof: inverse of two p.s.d matrices sum

If $\|\cdot\|_2$ denotes the maximum singular value, then it can be proved as follows.

The inequality in the original problem is equivalent to showing $$ \sigma_{\max}((A+B)^{-1})\le \sigma_{\max}(B^{-1}) \Leftrightarrow \frac{1}{\sigma_{\min}(A+B)}\le \frac{1}{\sigma_{\min}(B)} \Leftrightarrow \sigma_{\min}(B)\le\sigma_{\min}(A+B) $$ Since $A$ is PD, the last inequality above is easy to prove (consider the eigenvector for the minimum eigenvalue of $A+B$, then $\sigma_{\min}(A+B)=x^T(A+B)x\ge x^TBx\ge \sigma_{\min}(B))$.

Best Answer

Related Solutions

[Math] Counterexamples to the Matrix norm AM-GM inequality

[Math] Matrix norm inequality proof: inverse of two p.s.d matrices sum

Related Question