The problem has already been solved by a user who deleted his account so I ask a question regarding his answer. This was the posted problem:

Let $f:\Bbb R\to\Bbb R$ be continuous & periodic with prime period $\tau>0$. Prove: $\exists x_0\in\Bbb R$ s.t. : $$f\left(x_0+\frac{\tau}2\right)=f(x_0)$$

I've read the Bolzano-Weierstrass theorem & the Intermediate Value Theorem are equivalent in some sense:

If a continuous function is positive at some point on an interval and negative at another, it must be equal to zero somewhere.

His answer was accepted and applies only the tools I've heard of in the Analysis 1 lecture so far.

The constructed function:$$g(x):=f\left(x+\frac{\tau}2\right)-f(x)$$ The part I would like to understand: then $$\begin{aligned}g(0)g\left(\frac{\tau}2\right)&=\left(f\left(\frac{\tau}2\right)-f(0)\right)\left(f(\tau)-f\left(\frac{\tau}2\right)\right)\\&=\left(f\left(\frac{\tau}2\right)-f(\tau)\right)\left(f(\tau)-f\left(\frac{\tau}2\right)\right)\\&=-\left(f(\tau)-f\left(\frac{\tau}2\right)\right)^2\leq 0\end{aligned}$$ Did the $OP$ apply the quoted form of the W-B theorem? Can I generalize the statement to a segment $[a,a+\tau]$ as in many other posts?


Recently I've found a nice & rigorous Weierstrass theorem (regarding continuous functions on a closed interval/segment) statement. As some of my colleagues said our script isn't quite neat, I've decided to quote the author Svetozar Kurepa here in case if Croatian students ever see this post.

literature: prof. dr. Kurepa, S.: Math Analysis, part 2: One-variable functions page 31, theorem no. 4:

Let $f$ be a real function continuous on $I=[a,b]$

$(1)$ $f$ is bounded on $I$

$(2)$ $\exists m,M\in\Bbb R,\ x_m,x_M\in I$ s.t. $$m\leq f(x)\leq M\;(\forall x\in I)\ \ \&\ \ m=f(x_m),\;M=f(x_M)$$ i.e. function $\underline{\text{continuous}}$ on a segment reaches its minimum & and maximum

$(3)$ $\forall C\in\langle m,M\rangle\;\exists c\in\langle a,b\rangle$ s.t. $C=f(c)$ i.e., continuous function reaches every intermediate value on a segment.

$(4)$ $$[(f(a)<0;\land; > f(b)>0)\lor(f(a)>0\;\land\;f(b)<0)]\implies\;\exists c\in\langle a,b\rangle\;s.t.\;f(c)=0$$

$1^{\text{st}}$note: the proof is on the pages 350-351.

$2^{\text{nd}}$ note: for the proof of the first of the three statements it is crucial: Just like the image/range of a function on a segment in a codomain: $$R_f\in[f(a),f(b)]\subseteq\mathcal C_f,$$ any sequence $(x_n)$ contained in the preimage of a function (which is a segment) is bounded: $$(x_n)\in[a,b]\subseteq\mathcal D_f\implies\;(x_n)\;\text{is bounded}$$

By the author, the intermediate value theorem (teorem o međuvrijednosti) is one of the $4$ statements of the Weierstrass theorem given above.

  • 1
    $\begingroup$ Yes, the OP used the quoted theorem. Call that thing you quoted "The intermediate value theorem". Being "equivalent in some sense" to the Bolzano-Weierstrass theorem is no reason to simply call it "the Bolzano Weierstrass theorem". When you define $g$, show its variable ... $g(x_0) = \cdots$ $\endgroup$
    – GEdgar
    Commented Jan 24, 2020 at 19:14
  • $\begingroup$ @GEdgar, thank you for the response, I saw I wrote $x_0$ instead of $x$. $\endgroup$
    – PinkyWay
    Commented Jan 24, 2020 at 19:19
  • $\begingroup$ But you did not write $g(x)$ for your definition. $\endgroup$
    – GEdgar
    Commented Jan 24, 2020 at 19:34
  • 1
    $\begingroup$ It's a special case of universal chord theorem. You can replace $\tau/2$ with $\tau/n, n\in\mathbb {N} $ and result still holds. $\endgroup$
    – Paramanand Singh
    Commented Jan 25, 2020 at 4:08

1 Answer 1


Did the OP apply the quoted form of the W-B theorem?

Yes, he used the theorem you quoted. It is usually referred to as the intermediate value theorem.

A more detailed proof would go as follows.

Note that the function $g$ is continuous. By showing that $g(0) g(\frac{\tau}{2}) \leq 0$ one can conclude that $g$ changes signs inbetween $0$ and $\frac{\tau}{2}$. Thus, by the intermediate value theorem there exists $x_0 \in (0,\frac{\tau}{2})$ s.t. \begin{align} g(x_0) &= 0 \\ \implies f\left(x_0+\frac{\tau}{2}\right) &= f(x_0)\,. \end{align}

Can I generalize the statement to a segment $[a,a+\tau]$ as in many other posts?

Yes. The proof is almost the same as above, just replace the function $g$ by a shifted $g_a$ such as $$ g_a(x) := g(x-a)\,. $$


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