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Let $f$ be continuous real-valued function on $[0,1]$ and

\begin{align} F(x)=\max\{f(t):0\leq t\leq x\}. \end{align} I want to show that $F(x)$ is also continuous on $[0,1]$.

MY WORK

Let $\epsilon> 0$ be given and $x_0\in [0,1].$ Since f is continuous at $x_0\in [0,1],$ then $\forall x\in [0,1]$ with $|x-x_0|<\delta,$ it implies $|f(x)-f(x_0)|<\epsilon.$

Also, \begin{align} |f(t)-f(x_0)|<\epsilon, \text{whenever}\; |t-x_0|<\delta,\;\forall\; t\in[0,x]\end{align}

Taking max over $t\in[0,x]$, we have \begin{align} \max|f(t)-f(x_0)|<\epsilon, \text{whenever}\; |t-x_0|<\delta,\;\forall\; t\in[0,x]\end{align} \begin{align} |\max f(t)-\max f(x_0)|<\epsilon, \text{whenever}\; \max|t-x_0|<\delta,\;\forall\; t\in[0,x]\end{align} \begin{align} |F(x)-f(x_0)|<\epsilon, \text{whenever}\; |x-x_0|<\delta\end{align} which implies that $F(x)$ is continuous on $[0,1].$

I am very skeptical about this proof of mine. Please, is this proof correct? If no, a better proof is desired. Thanks!

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  • $\begingroup$ There are issues in the last three "equations" you wrote. More precisely in the implication to move from one to the next one. A very important principle in maths. Explain (at least to yourself) what theorem, axiom... you use when you move from one affirmation to the next one. $\endgroup$
    – user532133
    Sep 7, 2018 at 8:51

2 Answers 2

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Your proof is wring because you cannot take maximum over $t \in [0,x]$ in an inequality which is valid only for $|x-t| <\delta$. Here are some hints for a correct proof. Verify that $|F(x)-F(y)| \leq \max \{|f(t)-f(s)|:x\leq t \leq y, x\leq t \leq y\}$ for $x<y$ $\,\,$ ($\dagger$) . Once you do this you can use the fact that the continuous function $f$ on $[0,1]$ is uniformly continuous to prove that $F$ is also uniformly continuous. Hint for ($\dagger$) : if $x\leq t \leq y$ then $f(t)=f(x)+[f(t)-f(x)]$.

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  • $\begingroup$ Sir @Kavi Rama Murthy! Please, I am not getting how you got this $|F(x)-F(y)| \leq \max \{|f(t)-f(s)|:x\leq t \leq y, x\leq t \leq y\}$ for $x<y$ $\,\,$ ($\dagger$) Please, can you show me $\endgroup$
    – user586973
    Sep 7, 2018 at 9:04
  • $\begingroup$ In the hint I have given take max over $t$. You will get $\max_{x\leq t \leq y} f(t) \leq f(x)+\max_{x\leq t \leq y} |f(t)-f(x)|$. Note that $f(x) \leq F(x)$ and $\max_{0\leq t \leq x} f(t) \leq F(x)$. Hence $F(y) \leq F(x) +\max_{x\leq t \leq y} |f(t)-f(x)|$. $\endgroup$ Sep 7, 2018 at 9:12
  • $\begingroup$ it seems $\max_{x\leq t \leq y}f(t)\leq F(y)$. How you get left hand $F(y)\leq$? $\endgroup$
    – Riemann
    Sep 7, 2018 at 9:32
  • $\begingroup$ If max over $[0,x]$ and max over $[x,y]$ are both $\leq M$ for some number $M$ then max over $[0,y]$ is $\leq M$. $\endgroup$ Sep 7, 2018 at 9:35
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    $\begingroup$ First, $f(t)=f(x)+[f(t)-f(x)]\leq f(x)+|f(t)-f(x)|.$ So you can get $\max_{x\leq t \leq y} f(t) \leq f(x)+\max_{x\leq t \leq y} |f(t)-f(x)|$. Using $f(x) \leq F(x)$, we can get $\max_{x\leq t \leq y} f(t) \leq F(x)+\max_{x\leq t \leq y} |f(t)-f(x)|$. Next, both $\max_{0\leq t \leq x} f(t)(=F(x))\leq F(x)+\max_{x\leq t \leq y} |f(t)-f(x)|$ and $\max_{x\leq t \leq y} f(t) \leq F(x)+\max_{x\leq t \leq y} |f(t)-f(x)|$, you will get $\max_{0\leq t \leq x} f(t)=F(y)\leq F(x)+\max_{x\leq t \leq y} |f(t)-f(x)|.$ $\endgroup$
    – Riemann
    Sep 7, 2018 at 10:14
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Hint: by definition your function is monotone non-decreasing, hence it is enough to prove that it has no jumps.

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