$3f(2x + 1) = f(x) + 5x$. Find continuous $f$.

Question

Let some continuous $f : \mathbb R \rightarrow \mathbb R$ satisfies $3f(2x + 1) = f(x) + 5x$. Find all $f$.

I don’t know how to deal with continuity of functions.

Help please. Thank you very much.

Answer 1

By plugging in $x=-1$, we find $$f(-1)=-\frac 52.$$

With $g(x):=f(x-1)-x+\frac52$ (or $f(x)=g(x+1)+x-\frac32$), we have $$ 3\bigl(g(2x+2)+2x-\tfrac12\bigr)=g(x+1)+6x-\frac{3}2$$ or after substituting $x\leftarrow x-1$ and simplifying, $$ 3g(2x)=g(x).$$ Then with $h(x):=g(2^x)$, we find $$ \tag13h(x+1)=h(x).$$ Assume for some $a,b\in \Bbb R$ with $b\ge a+1$, we have a function $h_{a,b}\colon [a,b]\to\Bbb R$ that is continuous and such that $(1)$ holds for $h_{a,b}$ and all $x\in[a,b-1]$. Then for $a'=a-1$, $b'=b+1$, there exists a unique function $h_{a',b'}\colon[a',b']\to\Bbb R$ such that $h_{a',b'}|_{[a,b]}=h_{a,b}$ and $(1)$ holds for $h_{a',b'}$ for all $x\in [a',b'-1]$. Indeed, wee only need to use $(1)$ to extend the function to the right by $h_{a',b'}(x)=\frac13(h_{a,b}(x-1)$ for $b<x\le b'$, and to the left by $h_{a',b'}(x)=3h_{a,b}(x+1)$ for $a'\le x<a$. Then clearly $h_{a',b'}$ is continuous when restricted to $[a,b]$, as well as on $(b,b']$ and on $[a',a)$. But as $(1)$ already holds for $h_{a,b}$ with $x=a$ and $x=b-1$, respectively, we see that in fact $h_{a',b'}$ is continuous also at $a$ and $b$, i.e., on all of $[a',b']$.

Then by induction, for every continous $h_{0,1}\colon[0,1]\to\Bbb R$ with $3h_{0,1}(1)=h_{0,1}(0)$, we find a unique function $h\colon\Bbb R\to\Bbb R$with $h|_{[0,1]}=h_{0,1}$ and such that $(1)$ holds for all $x\in\Bbb R$; and this $h$ is continuous.

Each such $h$ allows us to define $g\colon(0,\infty)\to\Bbb R$ as $g(x)=h(\log_2x)$ and ultimately $f\colon(-1,\infty)\to\Bbb R$ as $f(x)=g(x+1)+x-\frac32$ to obtain a function that is continuous in its domain and obeys the functional equation in its domain.

We want $f$ not only continuous on $(-1,\infty)$ but on all of $\Bbb R$, in particular at $-1$. So we need $$ -\frac 52=f(-1)=\lim_{x\to -1^+}f(x)=\lim_{x\to0^+}g(x)-\frac52=\lim_{x\to-\infty}h(x-\frac 52).$$ As $h(x-n)=3^nh(x)$, the latter limit can only exist if $h(x)=0$ for all $x$. So all the variational choice we envisioned for $h$ (and $g$ and $f$) is made void by the demand for continuity at $x=-1$. And with $h(x)=0$ we find $g(x)=0$ for all $x>0$ and finally $f(x)=x-\frac 32$ for all $x>-1$.

To investigate $f$ on $(-\infty,-1)$, we can redo the argument above, but with $h(x)=g(-2^x)$. We find that the only continuous solution is $$f(x)=x-\frac32. $$ (In fact, this is even the only solution if we forget continuity and instead merely demand that $f$ is bounded in some neighbourhood of $-1$)

Answer 2

Hints:

Shift $x$ by a suitable constant to get rid of the constant $+1$ and get the form $3g(2y)=g(y)+\cdots$. Then write the argument as a power of $2$ and you'll get a linear recurrence $h(z+1)=h(z)+\cdots$.

After solving for $h,g$ then $f$, discuss continuity.

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With $x=2^t+1$,

$$f(x)=f(2^t+1)=g(2^t)=h(t)\to f(2x+1)=g(2\,2^t)=h(t+1)$$ and

$$h(t+1)=\frac13h(t)+\frac53(2^t+1).$$