continuous (on 3, 4 and 5) f is constant, if $f(x+2)+f(4x)=f(2x+1)+f(2x+2),\forall x\in\mathbb{R}$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a function that is continuous on 3, 4 and 5, such that $f(x+2)+f(4x)=f(2x+1)+f(2x+2),\forall x\in\mathbb{R}$. Show that f is constant.

I don't know what to do here!! I think that we may be able to prove that f is continuous on $\mathbb{R}$, but then what? Also, I don't have any school knowledge of sequences, but only some calculus, including integrals, derivatives, IVT, MVT, EVT, Rolle, Fermat, Darboux, limits and such things.

Any hint of a proof that doesn't use Taylor or sequences?

Answer 1

Denote $g(x) = f(2x) - f(2x-2)$. Then given condition implies that $g(x) = g(2x-3)$ and $g$ is continuous at $x_0=3$. Now suppose, there exists $y<3$ such that $g(y)\neq g(3)$. Set $a$ = $\limsup_{g(y)<g(3), y<3} y$. Because $g$ is continuous at $3$, such $a$ exists. However, $g(3-\epsilon) = g(3-2\epsilon)$ for $\epsilon>0$. Contradiction. So $g$ is constant. Now only you need to show that $g(x) = 0$ for all $x\in\Bbb{R}$.

Answer 2

The statement is false. Continuity at $2$ is necessary and sufficient for the function to be constant. Continuity at all $x \neq 2$, which is more than what is assumed in the problem, is not sufficient for the function to be constant.

The functional equation is

$(T^2 + D^2) f = (TD + DT) f$,

or equivalently, $(T - D)^2 f = 0$, where

• $D$ is the doubling operator $f(x) \to f(2x)$ and

• $T$ the translation operator $f(x) \to f(x+1)$.

First we solve $(T-D)g = 0$. This says $g(x+1) = g(2x)$ or $g(2+t)=g(2+2t)$. The solution is any function $g$ invariant under the transformation $2+t \to 2+2t$. The only solutions continuous at $2$ are $g = $ constant. There are nonconstant solutions continuous everywhere except at $2$, which shows that the conclusion of the problem is incorrect even for the stronger functional equation $f(x)=f(2x-2)$ that implies the 4-term equation in the problem.

Next, to solve the 4-term functional equation, we want $g = (T-D)f$ to be a solution of $(T-D)g = 0$. The functions $f$ making that true are exactly the functions linear in $n$ along geometric progressions $(2 + 2^n a)$. The linear function can be any $P(a) + nQ(a)$ where $P$ and $Q$ are arbitrary functions of $a \in [1,2)$.

Continuity at points other than $x=2$ is not enough to force the function to be constant, or to have $Q=0$, or to have $P$ or $Q$ be constant.