If $f$ is differentiable on a closed bounded interval, is $f$ bounded?

Question

I am working on this problem:

I want to prove that when $f$ is differentiable on a closed interval $[x_1,x_2]$, where $x_2>x_1$, $f$ is bounded.

At first I was thinking that I could prove (by contradiction) that $f$ had to be bounded because $\infty$ cannot be on a closed interval. And since $f$ is on a closed interval, it can't be unbounded. But I think we're supposed to use the mean value theorem.

So, I know that $f$ is bounded on $I$ when there exists a constant K such that $|f(x)|\leq K$ for all $x \epsilon I$. Since $f$ is differentiable, it is continuous. Vi can thus use the mean value theorem. We manipulte the formula a bit and get: $f'(c)(x_2-x_1)=f(x_2)-f(x_1)$, since $f$ is bounded at $|f(x)|\leq K$, $|f'(c)||x_2-x_1|=|f(x_2)-f(x_1)|$

Now, I'm trying to get to a point where I can prove that $f\leq K$, probably using the triangle inequality? Also that I could solve the MVT-expression for $f(x)$ when the $|MVT|\leq K$? Am I on to something?

Answer 1

That a function $f$ defined and continuous on a closed and bounded interval $[a,b]$ is bounded is ultimately a consequence of the Bolzano-Weierstrass theorem. BWT states that any bounded sequence has a convergent subsequence.

You could use BWT indirectly to prove that $f$ is bounded. Assume that $f$ is unbounded. This implies that for any positive integer $n$ there is a sequence $(x_n)$ in $[a,b]$ such that $|f(x_n)| > n$ -- leading eventually to a contradiction.

On the other hand, you can use another consequence of BWT, that the closed and bounded interval $[a,b]$ is compact, implying that any open cover has a finite subcover. In this way you can salvage your proof. You have shown that for any point $x \in [a,b]$ there exists an interval $I_x = (x - \delta_x,x + \delta_x)$ where $f$ is locally bounded, i.e. $|f(y)| < M_x$ for all $y \in I_x$.

Since $[a,b] \subset \cup_{x \in [a,b]}I_x$, then compactness implies there are a finite number of points $x_1, x_2, \ldots , x_n \in [a,b]$ such that $[a,b] \subset \cup_{k=1}^nI_{x_k}.$ Take $M = \max(M_{x_1}, M_{x_2}, \ldots, M_{x_n}).$ For any $x \in [a,b]$ there exists $1 \leqslant j \leqslant n$ such that $x \in I_{x_j}$and $|f(x)| < M_{x_j} < M$.

Answer 2

edited

Ok so I tired to combine theorem 1 and 5 and collary 1.

$f(x)$ is differentiable on $[x_1,x_2]$, thus it is also continuous on $[x_1,x_2]$. Given that $f(x)$ is continuous, there exists a $\delta>0$ such that $f$ is bounded on $(x_0-\delta, x_0+\delta)$, which leads to $f(x)$ being bounded on the neighbourhood $x_0$ on $[x_1,x_2]$. Furthermore, if $f$ is bounded on $x_0$, $f$ itself is bounded on $[x_1,x_2]$. Which was to be proven. $\blacksquare$

Answer 3

Let $S$ be a compact space and let $T$ be any space . If $f:S\to T$ is continuous then the image $f''S=\{f(x):s\in S\}$ is a compact subspace of $T.$ In other words, the continuous image of a compact space is compact.

Proof:

Let $G$ be an open family in $T$ with $\cup G\supset f''S.$ Then $J=\{f^{-1}H:H\in G\}$ is an open family in $S$ because $f$ is continuous. Obviously $\cup J=S,$ and so we can take a finite $J^*\subset J$ with $\cup J^*=S.$

For each $j\in J^*$ take $\psi(j)\in H$ satisfying $f^{-1}\psi (j)=j.$ Then $\{\psi (j):j\in J^*\}$ is a finite subset of G, and is a cover of $f''S.$ QED.

If $f:[a,b]\to \mathbb R$ is differentiable then $f$ is continuous. And $[a,b]$ is compact. So $f''[a,b]$ is compact, hence it is bounded.

The notation $f''A=\{f(a):a\in A\}$ when $A$ is a subset of the domain of $f,$ is read "$f$ double-prime $A$".