A stupid question... if $f$ is continuous on $[a,b]$, then is it also bounded? If so, how do we write its upper bound? Is it $f(x)\leq f(b)$ for any $x \in [a,b]$?

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    $\begingroup$ Image of a compact set under a continuous map is compact, but the upper bound is not necessarily $f(b)$ $\endgroup$ – J. W. Tanner Mar 19 at 16:28
  • $\begingroup$ No it is not! Think about $\sin x$ in the interval $\left[ 0, \pi \right]$. $\endgroup$ – Aniruddha Deshmukh Mar 19 at 16:28
  • $\begingroup$ It is bounded, but not necessarily by $f(b)$. $\endgroup$ – Don Thousand Mar 19 at 16:28
  • $\begingroup$ Nope. Take the function $f(x)=b-x,x \in [a,b]$ where I assume $a < b.$ $\endgroup$ – Dbchatto67 Mar 19 at 16:28

Yes $f$ will be bounded on $[a,b]$ and moreover $f$ will achieve it's maximum on the interval $[a,b]$. $f$ being bounded on $[a,b]$ means that there exists $M > 0$ such that $|f(x)| \leq M$ for all $x \in [a,b]$. On the other hand, achieving its maximum on $[a,b]$ means that there exists $c \in [a,b]$ such that $f(x) \leq f(c)$ for all $x \in [a,b]$. However, there still may be a point $x \in [a,b]$ such that $|f(x)| > f(c)$. Note also that $c$ might not be $b$!

Let's now prove that $f$ is bounded on $[a,b]$. Otherwise, for every $M > 0$, we could find some point $x_M \in [a,b]$ such that $|f(x_M)| > M$. In particular, there exists a sequence $(x_n)$ in $[a,b]$ such that $$ |f(x_n)| \geq n, \quad \forall n \geq 1. $$ Because $[a,b]$ is bounded, $(x_n)$ has a subsequence $(x_{n_k})$ that converges to a point $x \in [a,b]$. Then, by continuity, we must have $$ f(x) = \lim_{k \to \infty} f(x_{n_k}) $$ whence $(f(x_{n_k}))$ is convergent and bounded. However, $|f(x_{n_k})| \geq n_k$ for all $k \in \mathbb{N}$ ensures that this sequence cannot be bounded. This gives us our contradiction.


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