# Constructive proof of boundedness of continuous functions

Consider the theorem for the continuous function:

Let $a<b$ be real numbers, and let $f:[a,b]\to{\bf R}$ be a function continuous on $[a,b]$. Then $f$ is a bounded function.

The proof in the classical textbook on real analysis uses the Heine-Borel theorem. It dose not say how to find the bound for $f$, but it show that having $f$ unbounded leads to a contradiction.

Here are my questions:

• Is there a direct [EDITED: constructive] proof for this theorem?

• More generally, can a theorem in mathematics always have a constructive proof? Or what kind of statements do not have any constructive proof, say, one has to use techniques such as "proof by contradiction" in order to prove it?

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Given the Gödel-Gentzen transformation you could argue that any theorem in mathematics (ZF(C)) can be proved constructively. You actually prove a classically equivalent statement (and it is not informative given that the final result is a negation) but... – gallais May 26 '11 at 7:15

There's a natural proof using sup. Consider the set $X=\{ x \in [a,b] : f \mbox{ is bounded in } [a,x] \}$. Then $X$ is non-empty and no $t<b$ is an upper bound for $X$. Hence $b=\sup X$ and $f$ is bounded in the whole interval. This argument is actually the same one used in Heine-Borel: for continuous functions in compact sets, locally bounded implies globally bounded. That $f$ is locally bounded comes from continuity and is used to prove that no $t<b$ is an upper bound for $X$.

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Is this supposed to be a proof? – Glen Wheeler May 26 '11 at 14:21
@Glen, yes. Or a sketch, rather. This proof is given in full in Spivak's Calculus, Theorem 7-2. – lhf May 26 '11 at 14:58

There is no a-priori bound on the function: consider the constant function $f = M$. Also, the maximum can be attained at any given point - just take any appropriate quadratic.

As for the second question, that really depends on your notion of "directly".

In constructive mathematics, I believe that a continuous function is supplied with some "modulus of continuity" which immediately implies (directly) boundedness. Whether one can prove intuitionistically that the $\epsilon-\delta$ definition of continuity implies boundedness - probably not, but ask an expert!

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As for your question about statements that can be proven "directly" or not, I find that this article by Tim Gowers and the resulting discussion in the comments is very interesting. It even addresses the Heine-Borel theorem.

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Let $\{x_n\}_{n=1}^{\infty}$ be a sequence of points such that $|f(x_n)|$ converges to $M = \sup_x |f(x)|$. You can pass to a convergent subsequence $\{x_{n_k}\}_{k=1}^{\infty}$ which converges to some $y \in [a,b]$. Then by continuity of $f(x)$ $$|f(y)| = \lim_{k \rightarrow \infty} |f(x_{n_k})| = M$$ So $f(y) = \pm M$; therefore $M$ is finite and $|f(y)|$ achieves the value $M$. The proof of the existence of convergent subsequences as in the proof of the Bolzano-Weierstrass theorem, using divisions into halves and so on, is not by contradiction and is reasonably constructive.

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I don't see how this is any more direct than the proof outlined in the OP. – Qiaochu Yuan May 26 '11 at 15:42
it's not a proof by contradiction, the existence of a convergent subsequence is not proven in that way. This is what the poster asked for. – Zarrax May 26 '11 at 16:00
Also, the usual proof of the existence of convergent subsequences; i.e. repeated dividing into halves as in the Heine-Borel theorem, is appropriately constructive given the generality of the question. – Zarrax May 26 '11 at 16:13
@Qiaochu, @Zarrax: I agree with Qiaochu here, this proof doesn't completely work for me. It seems like you are saying this: "Let $M$ be the supremum. Then by sequence argument, and continuity, there exists $y$ such that $f(y)=M$. But $f(y)\in\mathbb{R}$ so $M<\infty$." Worded this way, it doesn't completely make sense, and is just a proof by contradiction in disguise. I feel the $g(x)$ you made is just to disguise this further. – Eric Naslund May 26 '11 at 16:56
This is exactly the proof in Tim Gower's article, few posts above ;) And he makes an interesting point about it: how do you prove that $x_n$ exists? If $M$ is finite, there is nothing to prove, but if $M$ is infinite, the existence of $x_n$ is based on the assumption that $M$ is infinite, how is this different than contradiction? – N. S. May 26 '11 at 18:20

Would this be considered a direct proof?

Pick $x_n$ a sequence which is dense in $[0,1]$. Let $y_n =f(x_n)$ and $z_n =\max_{1 \leq i \leq n} y_i$.

Then $z_n$ is increasing, and thus has a limit.

Let $A:= \{ m | z_m > z_{m-1} \}$. If $A$ is finite, it has a maximum $k$ and since $y_n \leq z_k$ for each $n$, it follows that $z_m$ is the maximum of $f(x)$.

If $A$ is infinite, then order its elements incresingly $n_1 < n_2 < ..< n_k< ...$. Then by the definition of $A$, the sequence $y_{n_k}$ is increasing and

$$y_{n_k}=z_{n_k}= \max_{1 \leq i \leq n_k} y_i (*) \,.$$

Finally pick a convergent subsequence $x_{m_l}$ of $x_{n_k}$. Let $a$ be the limit of this.

By continuity

$$\lim_{l \to \infty} y_{m_l}= f(a) \,.$$

Since $y_{m_l}$ is a subsequence of $y_{n_k}$, and $y_{n_k}$ is increasing we get

$$\lim_{k \to \infty} y_{n_k}= f(a) \,.$$

But then, since $y_{n_k}$ is increasing, we get from $(*)$ that $y_n \leq f(a)$ for all $n$. Now the density of $\{ x_n \}$ completes the proof.

Note We actually used Heine-Borel theorem in the proof, when we picked a convergent subsequence of $x_{k_n}$ (unless of course one defines compactness that way). No matter what proof one tries, since the same theoremfails for $(0,1)$, somehow compactness is bounded to come in play....

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