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So I have this function $f : \mathbb{R} \to \mathbb{R}$ that is continuous and I have $a\in\mathbb{R}$.

I have to prove that exists an $x_{0}\in\mathbb{R}$ such that this works:

$$f(|x_{0}+a|) = f(|x_{0}|)$$

So I started with creating a new function:

$$F(x)=f(|x_{0}+a|)-f(|x_{0}|)$$

but I am stuck with choosing the boundaries of domain and codomain of $F$, so I can't really find the zero of this function. If I started the wrong way correct me please. Any help would be appreciated.

Important: this is to be solved only with help of some basic theorems of functions continuity. This is from 1st semester of calculus.

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    $\begingroup$ As written, you don't need continuity of $f$. Such an $x_0$ exists for every function $f\colon \mathbb{R}\to \mathbb{R}$, however wild. $\endgroup$
    – Daniel Fischer
    Commented Jun 28, 2015 at 12:14
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    $\begingroup$ ^Rather, choose $x_0=-\frac a2$. Then $|x_0+a|=|x_0|$ and so you are through. $\endgroup$
    – Chaitanya Tappu
    Commented Jun 28, 2015 at 12:15
  • $\begingroup$ DanielFischer & cmtappu96 : This is indeed surprisingly easy.. I really thought one would somehow have to use Mean Value Theorem etc. $\endgroup$
    – Imago
    Commented Jun 28, 2015 at 12:18
  • $\begingroup$ More likely the intermediate value theorem, since the MVT requires derivatives. :) $\endgroup$
    – John Hughes
    Commented Jun 28, 2015 at 12:22
  • $\begingroup$ thankyou, i actually didn't thought of just using intermediate value. $\endgroup$
    – MathIsTheWayOfLife
    Commented Jun 28, 2015 at 12:35

1 Answer 1

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$F(x)=f(|x+a|)-f(|x|)$

We have 3 cases.

$1)$ $F(0) = 0 $

$2)$ $F(0) = f(a)-f(0) > 0 \implies F(-a) = f(0)-f(a) < 0$

$3)$ $F(0) = f(a)-f(0) < 0 \implies F(-a) = f(0)-f(a) > 0$

By composition of continuous functions F is continuous so we apply IVT and find $F(x_0)=0$ for some $ x_0 \in [-a,0] $

$F(x_0)=0 \implies f(|x_0+a|=f(|x_0|)$

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  • $\begingroup$ Thank you, but i guess it is enough to take an $x_{0}=-\frac{a}{2}$ which works in general,for any a, because of absolute value, it's easier to show. Thanks to one of the comments on the question. But your answer is also appreciated and correct :) $\endgroup$
    – MathIsTheWayOfLife
    Commented Jun 28, 2015 at 13:50

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