Conditions on taking supremum over $t$ for $|f(t)|$?

Question

Let $f: [0,T]\times X^{n} \to X$ is a continuous function given by $f \big(t, \mu_{1}(t), \mu_{2}(t), \dots,\mu_{n}(t) \big)$, $t\in [0,T]$. Here $X$ is a Banach space, and $\mu_{i}\in C([0,T],X),~ \text{for}~(i=1,2,\dots,n)$.If \begin{equation*} \big|f \big(t, \mu_{1}(t), \mu_{2}(t), \dots,\mu_{n}(t) \big)-f \big(t, \mu_{1}^{*}(t), \mu_{2}^{*}(t), \dots,\mu_{n}^{*}(t) \big)\big|\leq \sum_{i=1}^{n}|\mu_{i}-\mu_{i}^{*}| \end{equation*} Can we take supremum over $t$, and get \begin{equation*} \big\|f \big(t, \mu_{1}(t), \mu_{2}(t), \dots,\mu_{n}(t) \big)-f \big(t, \mu_{1}^{*}(t), \mu_{2}^{*}(t), \dots,\mu_{n}^{*}(t) \big)\big\|\leq \sum_{i=1}^{n}\|\mu_{i}-\mu_{i}^{*}\| \end{equation*} where $\|f(t)\|=\sup\{|f(t)|:t\in [0,T]\}$? If not then what are the additional conditions?

Answer 1

Let's simplify the notation first. Define $g_{i}(t)$ as the absolute difference between $\mu_{i}(t)$ and its corresponding value $\mu_{i}(t)^{*}$, i.e., $g_{i}(t)=|\mu_{i}(t)-\mu_{i}(t)^{*}|$. This holds for all $t$ in the interval $[0,T]$. It's clear that $g_{i}(t)$ is always less than or equal to the supremum of $g_{i}(t)$ over the interval $[0,T]$, denoted as $||\mu_{i}-\mu_{i}^{*}||$. Hence, we have $g_{i}(t)\leq||\mu_{i}-\mu_{i}^{*}||$ for all $t$ in $[0,T]$. As a result, the absolute difference between the functions $F(t)$ and $F^{*}(t)$ is bounded by the sum of $g_{i}(t)$ over all $i$, which is further bounded by the sum of $||\mu_{i}-\mu_{i}^{*}||$ over all $i$. Mathematically, this can be expressed as: $$|F(t)-F^{*}(t)|\leq\sum_{i=1}^{n}g_{i}(t)\leq\sum_{i=1}^{n}||\mu_{i}-\mu_{i}^{*}||.$$ This inequality holds for all $t$ in $[0,T]$. If we take the supremum over $t$ on both sides (noting that the right-hand side does not depend on $t$), we obtain: $$||F-F^{*}||\leq\sum_{i=1}^{n}||\mu_{i}-\mu_{i}^{*}||.$$ Here, $F$ and $F^{*}$ represent your respective functions.