Continuity of the norm of a matrix Lets say that $Q(t)$ is a $n\times n$ matrix that depends on $t\in\mathbb{R}$, and define the matrix $P(t)$ as 
\begin{align*}
P(t) = \int_0^t Q(u)\,du,
\end{align*}
where the integral is taken entry-wise and I assume that the entries of $Q(t)$ are integrable for every $t$. In that sense, the entries of $P(t)$ are continuous. But, what can I say about the continuity of the function $f(t)$ defined as 
\begin{align*}
f(t) = \lvert\lvert P(t) \rvert\rvert\;?
\end{align*}
where $||\cdot||$ is the spectral norm (or operator norm) defined as $\lvert\lvert A \rvert\rvert=\sup_{x\neq 0}\lvert\lvert Ax \rvert\rvert_2/\lvert\lvert x \rvert\rvert_2$. In particular, if $Q(t)$ is symmetric with rank 1, I can write $Q(t) = \lambda_tv^{T}_tv_t$, where $(\lambda_t,v_t)$ is the nonzero eigenpair of $Q(t)$. Then $f(t)$ is 
 \begin{align*}
f(t) = \lvert\lvert \int_0^t \lambda_u v^{T}_uv_u\,du \rvert\rvert\leq \int_0^t \lvert\lvert\lambda_u v^{T}_uv_u\rvert\rvert\,du = \int_0^t|\lambda_u|\,du.
\end{align*}
I'm stuck here. I don't know the answer to the question. I'll appreciate comments and clues...
 A: As mentioned in the comment above, there is only one norm on a finite-dimensional $\mathbf{R}$-vector space, up to equivalence. I'll make this a bit more precise.
A (non-degenerate) norm on an $\mathbf{R}$-vector space $V$ is a function $v \mapsto |v|$ from $V$ to $\mathbf{R}_{\geq 0}$ with the properties:


*

*$|av|=|a| |v|$ for all $a \in \mathbf{R}$ and $v \in V$,

*$|v+w| \leq |v|+|w|$ for all $v,w \in V$, and

*$|v|=0 \ \iff \ v=0$.
Two norms $|v|_1$ and $|v|_2$ are equivalent if there are $c_1, c_2 \in \mathbf{R}_{>0}$ such that
$$|v|_1 \leq c_1 |v|_2 \quad \text{and} \quad |v|_2 \leq c_2 |v|_1 \quad \hbox{for all $v \in V$.}$$ For a finite dimensional $\mathbf{R}$-vector space, any two norms are equivalent (for instance, one can prove that any norm on $\mathbf{R}^n$ is equivalent to the max norm given by the absolute value of the largest coordinate). As a consequence, every norm on a finite-dimensional $\mathbf{R}$-vector space is continuous with respect to the topology induced by any isomorphism $V \cong \mathbf{R}^n$ of $\mathbf{R}$-vector spaces. Applying this to the vector space of matrices shows that the operator norm you have defined is continuous, and hence its composition with $t \mapsto P(t)$ is continuous.
