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Let $f(x)$ be a derivable function at $x=0$ & $f\left(\dfrac{x+y}{k}\right)=\dfrac{f(x)+f(y)}{k} \left(k\in R, k\ne 0,2\right)$. Show that $f(x)$ is either a zero or odd linear function.

My attempt is as follows:-

Putting $x=0,y=0$ in the functional equation

$$f(0)=\dfrac{2f(0)}{k}$$ $$f(0)(k-2)=0$$ $$f(0)=0 \text { as it is given $k \ne 2$ }$$

Replace $y=-x$ in the functional equation.

$$f(0)=\dfrac{f(x)+f(-x)}{k}$$ $$f(x)+f(-x)=0$$ $$f(-x)=-f(x)$$

Thus $f(x)$ is definitely an odd function.

As $f(x)$ is a derivable function at $x=0$ then following limit should exist:-

$$\lim_{h\to 0}\dfrac{f(h)-f(0)}{h}$$ $$\lim_{h\to 0}\dfrac{f(h)}{h}$$

Assuming $f(x)$ to be polynomial function.

If $f(x)$ is a linear function, then limit should be non-zero because if its a polynomial of degree greater than $1$, then limit would be zero.

Like if $f(x)=ax^2+bx+c$

Applying L' Hospital rule:- $$\lim_{h\to 0}(2ah+b)=0$$

If $f(x)$ is a constant function other than $0$, then $f(x)=-f(x)$ will not hold.

So we just have to prove that $$\lim_{h\to 0}\dfrac{f(h)}{h} \ne 0$$

Applying L' Hospital rule as we have $\dfrac{0}{0}$ form and function is derivable at $x=0$

$$\lim_{h\to 0}f'(h)=f'(0)$$

If we can just prove that $f'(0) \ne 0$, then we will be done.

I am not getting how to prove this fact.Any hints?

Martin Sleziak
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user3290550
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  • You can't prove $f'(0) \neq 0$ because that's one of the solutions. Assuming $f$ is a polynomial is also unnecessary. Try to prove $f$ is linear instead. Hint: Let $x=ka$ and $y=kb$. – Ninad Munshi Dec 21 '19 at 18:36
  • yeah, but any hints how to prove if its linear? – user3290550 Dec 21 '19 at 18:37

2 Answers2

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If $x=y$ then for $a=\frac{2}{k}$ we have $$f(ax)=af(x)$$

Now for $k=1$ $$f(x+y)=f(x)+f(y),\forall x,y$$

Since $f$ is continuous, $f$ is linear.

Indeed you can prove that $f(q)=bq,\forall q \in \Bbb{Q}$ and then you can proceed using density of rationals and continuity.

Martin Sleziak
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Marios Gretsas
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Let $k=1$. Then we have that

$$f\left(\frac{x+y}{k}\right) = \frac{f(x)+f(y)}{k} \implies f(x+y) = f(x) + f(y)$$

Now let $y = 0$ and let $k = \frac{1}{t}$. Then we have

$$f(tx) = tf(x)$$

For all $t \neq \frac{1}{2}$. This last case can be proven separately.

Ninad Munshi
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  • but we can't let k to be of particular value, we have to prove for all k, right? – user3290550 Dec 21 '19 at 18:45
  • @user3290550 there is no "prove for all $k$", you are thinking backwards. We have a property that applies for almost all $k$, so of course it holds for specific $k$. – Ninad Munshi Dec 21 '19 at 18:48
  • sorry, but I am not able to understand how u proved that function is linear, can you please explain in detail? – user3290550 Dec 21 '19 at 18:50
  • @user3290550 what is the definition of linearity...? – Ninad Munshi Dec 21 '19 at 18:52
  • how $f(x+y)=f(x)+f(y)$ is a linear function – user3290550 Dec 21 '19 at 18:52
  • I only know the definition of linearity in context of polynomials, $f(x)=ax+b$? – user3290550 Dec 21 '19 at 18:53
  • @user3290550 my advice to you would be to look up what is the definition of linearity. It is a very important core definition to know and it never hurts to brush up on one's linear algebra. – Ninad Munshi Dec 21 '19 at 18:54
  • I got this definition of linearity:- "Linearity is the property of a mathematical relationship or function which means that it can be graphically represented as a straight line", isn't this the definition of linear polynomial? – user3290550 Dec 21 '19 at 18:57