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Let f be a real valued function on $\mathbb{R}$. If for all real $x$,it satisfies $$ f(x) + 3f(1-x) = 5$$ Then show that f is a constant function.

I tried it like this but not sure whether it is true or not.

$$ f(x) + 3f(1-x) = 5\tag 1$$

replace $x$ with $1-x$, $$f(1-x) + 3f(x) = 5\tag 2$$

Then solving (1) and (2) we get $f(x) = \frac{5}{4}$ for all $x\in \mathbb{R}$. Hence $f$ is a constant function.

Noel Lundström
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Satish
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2 Answers2

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Yes, your proof works, but don’t write $x=1-x.$ That is only true if $x=\frac12.$

Instead, write:

Let $y=1-x.$ Then:$$f(y)+3f(1-y)=f(1-x)+3f(x)=5.$$

and then finish the proof as you’ve done.

I realize that is what you really meant.

You could alternatively say, informally “substitute $1-x$ for $x.$” But don’t use the equal sign ($=$).

Some people might accept:

Letting $x:=1-x,\dots$

But that is not normally used.

Thomas Andrews
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  • It works over any set of numbers, even in quaternions. Any ring where $$3u+v=5\u+3v=5$$ has only one solution. So, It does not work in a ring with characteristic $2.$ @Cretin2 – Thomas Andrews May 24 '21 at 18:26
  • Then were it possible to do a mapping from $\mathbb{F}_{2^p}$ to $[0;1]$ via binary coding, and then to $\mathbb{R}$ via a tan for example. Then since the function is not asked to be continuous this could be a solution ? – QuantumPotatoïd Jun 18 '21 at 19:49
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I tried $f(z)=A\cos(az-b)^2$, then $f(z)+3f(1-z)=A(g\cos(az)^2+h\sin(az)^2+k\sin(2az))$, where g,h,k are expressions in a,b. The equation implies $g=h$ and $k=0$ with the 2 unknowns a,b. Apparently there are complex solutions $b=\arcsin(\sqrt{\frac{3\pm i\sqrt{15}}{6}})$ and $a=\arcsin(\sqrt{\frac{4-2\sin(b)^2}{6}})+b$. However a shall vanish to be coherent with the other solution, I'll have to check if I madea mistake.

QuantumPotatoïd
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