Proof that $\frac{3-x^{4}}{8x^{2}}$ is the only solution to the functional equation $f(x) + 3f\left(\frac{1}{x}\right) = x^{2}$

Question

How can we prove that $f(x) + 3f\left(\frac{1}{x}\right) = x^{2}$ has only one solution and no other solution and that solution is $\frac{3-x^{4}}{8x^{2}}$.

I know that this is a linear function equation, but can't find any sources which say that all linear functional equation will have only one solution.

Answer 1

If in addition $g(x) + 3g(1/x) = x^2$ then $$f(x) - g(x) + 3f(1/x) - 3g(1/x) = 0.$$ Define $h = f-g$ so that $$h(x) = -3 h(1/x).$$ Now swap $x \mapsto 1/x$ to find $$h(1/x) = -3h(x)$$ so that $$h(x) = 9 h(x).$$ This forces $h = 0$ and thus $f=g$.

Answer 2

We always have $x \ne 0.$

From

$$(1) \quad f(x) + 3f(\frac{1}{x}) = x^{2}$$

we get

$$(2) \quad f( \frac{1}{x})+3f(x)=\frac{1}{x^2}.$$

If we add $(1)$ and $(2)$ we derive

$$(3) \quad f(x)+f(\frac{1}{x})= \frac{1}{4}(x^2+\frac{1}{x^2}).$$

$(3) - (1)$ shows that

$$(4) \quad -2f(\frac{1}{x})=\frac{1}{4}(x^2+\frac{1}{x^2})-x^2.$$

Now substitute $\frac{1}{x} \to x$ in $(4)$. Easy computations give now the desired result.