$f(x)$ is continuous on $[a,b]$. Now we define a new function $M(t)$, for every $t\in[a,b]$

$$M(t) = \sup_{a \leq x \leq t} f(x).$$

Prove formally that $M(t)$ is continuous on $[a,b]$. (sup = supremum)

$M(t)$ is monotonically increasing function, but I don't see how to continue from here.

Also I would like to know what are the standard strategy for proving a function is continuous on a given interval (not a point) except showing its a sum/product/composition of another continuous functions.

  • $\begingroup$ Continuity is a pointwise condition. So, in order to prove that a function is continuous on an interval, you have to prove that it's continuous at each point in the interval. $\endgroup$ – Sammy Black Feb 26 '14 at 8:48
  • $\begingroup$ Pick $t<t'$ and notice that $$\left| M(t)-M(t') \right| \leq \left| \sup_{t \leq x \leq t'} f(x) \right|.$$ Now use the continuity of $f$ at $t$ (or $t'$, it does not really matter). $\endgroup$ – Siminore Feb 26 '14 at 8:49
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    $\begingroup$ A monotone function can be discontinuous only if it has jumps. So it is a good starting point to assume the existence of jump discontinuity of $M$. Then you will reach the contradiction that $f$ is also discontinuous at points of discontinuity of $M$. $\endgroup$ – Sangchul Lee Feb 26 '14 at 8:57
  • $\begingroup$ @Siminore: The inequality you have mentioned seems to suggest that $M(t)$ is an additive function (something like the variation of $f$ on $[a, t]$, but it is not the case. Also your inequality does not let the difference $M(t) - M(t')$ tend to zero as $t' \to t$ $\endgroup$ – Paramanand Singh Feb 26 '14 at 11:18

Consider $a < t < b$ and let us analyze where the supremum $M(t)$ is attained in interval $[a, t]$. If it is attained at an interior point in $(a, t)$ then it is clear that $f(t) < M(t)$ and by continuity all values of $f(x)$ near $t$ are also less than $M(t)$. It follows that $M(t + h) = M(t)$ for all small values of $h$. So that $M(t)$ is continuous. On the other hand if $M(t)$ is attained at point $t$ so that $f(t) = M(t)$ then by continuity all values of $f$ near $t$ are close to $M(t)$. Thus for any given $\epsilon > 0$ we have a $\delta > 0$ such that $$f(t) - \epsilon < f(x) < f(t) + \epsilon$$ for $|x - t| < \delta$. This means that $$M(t) - \epsilon = f(t) - \epsilon \leq \sup_{x \in (t - \delta, t + \delta)}f(x) \leq f(t) + \epsilon = M(t) + \epsilon$$ It now follows that $|M(t + h) - M(t)| \leq \epsilon$ for all $|h| < \delta$ and hence $M(t)$ is continuous at point $t$. The same proof can be adapted easily for $t = a, t = b$.

  • $\begingroup$ thanks . its a nice proof a lot shorter from what I've seen. $\endgroup$ – Boris Morozov Feb 26 '14 at 11:21
  • $\begingroup$ @BorisMorozov: Most proofs in introductory calculus and analysis are simple and short provided you understand the meaning of various inequalities involved. By the way thanks to you too for the compliments. $\endgroup$ – Paramanand Singh Feb 26 '14 at 11:31

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