# bound of $f$ on the interval $[a,b]$

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 $$x \in [a,b]$$?

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