Running Maximum of a Continuous Function

Detailed Solution

For each $x\in[0,1]$, the maximum in $$ M(x)=\max_{0\le t\le x}f(t) $$ exists because $f$ is continuous and $[0,x]$ is compact. First, $M$ is necessarily non-decreasing. If $0\le x_1<x_2\le1$, then $$ [0,x_1]\subset[0,x_2]. $$ Therefore, $$ M(x_1)=\max_{0\le t\le x_1}f(t) \le \max_{0\le t\le x_2}f(t)=M(x_2). $$ Hence $$ \boxed{M(x_1)\le M(x_2)}, $$ so $M$ is non-decreasing. It remains to prove continuity. Since $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous. Let $\varepsilon>0$. Choose $\delta>0$ such that $$ |u-v|<\delta \quad\Longrightarrow\quad |f(u)-f(v)|<\varepsilon. $$ Take $x,y\in[0,1]$ with $x<y$ and $y-x<\delta$. Since $f$ is continuous, there exists $t_y\in[0,y]$ such that $$ M(y)=f(t_y). $$ If $t_y\le x$, then $$ M(y)=f(t_y)\le M(x). $$ But $M$ is non-decreasing, so $M(y)\ge M(x)$. Thus $$ M(y)=M(x). $$ If $t_y>x$, then $$ |t_y-x|\le y-x<\delta. $$ By uniform continuity, $$ |f(t_y)-f(x)|<\varepsilon. $$ Since $f(x)\le M(x)$, $$ M(y)=f(t_y)<f(x)+\varepsilon \le M(x)+\varepsilon. $$ Together with $M(y)\ge M(x)$, we obtain $$ 0\le M(y)-M(x)<\varepsilon. $$ Thus $$ |M(y)-M(x)|<\varepsilon $$ whenever $|y-x|<\delta$. Therefore $M$ is continuous on $[0,1]$. Hence $$ \boxed{M\text{ is continuous and non-decreasing on }[0,1].} $$ So the correct answer is $$ \boxed{\mathrm{B}}. $$ Why A is false: Continuity of $f$ does not imply differentiability of $M$. For example, let $$ f(x)=1-2\left|x-\frac12\right|. $$ Then $f$ is continuous, but $$ M(x)= \begin{cases} 2x,&0\le x\le\frac12,\\[2mm] 1,&\frac12\le x\le1, \end{cases} $$ which is not differentiable at $x=\frac12$. Why C is false: The preceding argument shows that continuity of $f$ forces continuity of its running maximum $M$. Therefore $M$ cannot have jump discontinuities. Why D is false: Even for continuous $f$, the running maximum need not be differentiable. The example above demonstrates this.