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We can say that function $f(x), x \in \mathbb{R}$ is symmetric about a point $x = x_0$ if $f(x_0 - x) = f(x_0 + x)$ for all $x$ for which $f$ is defined. Now let's try to find a function which is symmetric about every point. It is easy to see that functions like $f(x) = const$ are such functions.

Is it possible to prove that there are no other functions that are symmetric about every point? One can prove this for differentiable functions by demostrating that it has to be $f'(x) = 0$ for a function to satisfy the condition. But what about non differentiable functions defined $\forall x \in \mathbb{R}$?

scrutari
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  • Any points $x,y$ are symmetric about their midpoint $\frac{x+y}2$. – Myunghyun Song Feb 16 '19 at 18:03
  • The question is about functions $f(x)$ of $1$ argument. Symmetry is related to geometric symmetry of the curve defined by the function with respect to vertical line $x = x_0$. – scrutari Feb 16 '19 at 18:08
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    Note that for $x_0 =\frac{x+y}2$ and $h=\frac{y-x}2$, $$f(y) = f(x_0+h) =f(x_0-h) = f(x)$$ for every $x,y$. – Myunghyun Song Feb 16 '19 at 18:10
  • Sorry @Song, I don't see how your comment is related to my question. $\frac{x + y}{2}$ is not function of $x$. – scrutari Feb 16 '19 at 18:13

2 Answers2

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The following is a related question that may be useful, at least for the references provided:

Does $\lim_{h\rightarrow 0}\ [f(x+h)-f(x-h)]=0$ imply that $f$ is continuous?

For simplicity, let's only consider functions $f:{\mathbb R} \rightarrow {\mathbb R}.$ As shown by @Song in the comments, if a function is symmetric about $x_0,$ then the function is constant. To see why, assume $f$ is symmetric about each point and choose $a \in \mathbb R$ and $b \in {\mathbb R}.$ The result follows if we can show that $f(a) = f(b).$ By symmetry at $x_0 = \frac{a+b}{2}$ with the choice $x = \frac{-a+b}{2}$ we have

$$f(x_0 - x) \; = \; f\left(\frac{a+b}{2} \, - \, \frac{-a+b}{2}\right) \; = \; f\left(\frac{a+b}{2} \, + \, \frac{-a+b}{2}\right) \; = \; f(x_0 + x) $$

Since $f(x_0 - x) = f(a)$ and $f(x_0 + x) = f(b),$ the displayed equalities above imply $f(a) = f(b),$ which completes the proof.

Let $\mathscr E_f$ denote the set of points of symmetry of a function. We've just shown that if $\mathscr E_f = {\mathbb R},$ then $f$ is constant. A natural question is, given some notion of "big" and some notion of "bad", does there exist a "bad" function $f$ such that $\mathscr E_f$ is "big"? Questions of this type are dealt with in Section 6.2.1 (pp. 214-218) of Symmetric Properties of Real Functions by Brian S. Thomson (1994). Among other results there, Corollary 6.18 (p. 218) states that there exists an everywhere discontinuous and non-Lebesgue measurable function $f$ in which $\mathscr E_f$ is such that for each open interval $(a,b),$ the set $\mathscr E_f \cap (a,b)$ has outer Lebesgue measure $b-a$ and is of the second (Baire) category. Incidentally, we can't simplify this statement by saying that $\mathbb R - \mathscr E_f$ has Lebesgue measure zero (or is a first category set) due to situations like those discussed here.

Another natural question is, given some notion of "small" and some notion of "bad", does there exist a "bad" function $f$ such that, for each $x_0 \in {\mathbb R},$ the set of points $x$ such that $f(x_0 - x) \neq f(x_0 + x)$ is "small"? Among other results, Sierpinski showed in 1936 that assuming the continuum hypothesis, we can have "non-Lebesgue measurable" for "bad" and "at most countable" for "small" (see Theorem 6.20 on p. 218 of Thomson's book).

Finally (at least for my answer), still another natural question is what if we keep the requirement "for all $x_0 \in \mathbb R$" but replace the requirement "for all $x \in \mathbb R$" with "for all $x$ in some open interval containing $x_0$" (allowing for the possibility that the size of these open intervals can vary with $x_0).$ In this case, one can show that there exists a closed and at most countable set $E$ such that $f$ is constant on $\mathbb R - E.$ Two proofs (at least) of this result are given in Thomson's book, one on p. 49 and the other on p. 223.

Dave L. Renfro
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This is very simple: given points $a,b\in\Bbb R$, we know that $f$ is symmetric about their mid-point $x_0=(a+b)/2$, so

$$f(a)=f(x_0-(b-a)/2)=f(x_0+(b-a)/2)=f(b)$$

Hence $f$ is constant.

TonyK
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