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A continuous function satisfies the condition $$f(x)=\frac{1}{t}\int_{0}^{t}[f(x+y)-f(y)]dy\quad\forall x\in\mathbb{R}, \forall t>0$$

We need to show $f(x)=cx$ for some constant $c$, well from the data we see $f(0)=0$ and I am unable to show $f(x+y)=f(x)+f(y)$ as then I can claim $f(x)=cx$ right? Would any one give me hint for that?

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up vote 3 down vote accepted

Fix $x$ and let $g(t) = tf(x)$. Then on the one hand $g'(t) = f(x)$ and on the other (fundamental theorem of calculus) $g'(t) = f(x + t) - f(t)$. Equating the right hand sides gives what you want.

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fine! but how did you come to know you need take $g(t)=tf(x)$, or it is just your intuition? – Un Chien Andalou Apr 28 '12 at 16:57
I wanted to use the second fundamental theorem of calculus, because I saw an integral with a variable in the bounds, so I needed to differentiate something with respect to $t$. You could also directly differentiate both sides of the equality with respect to $t$ and use the product rule on the RHS (the LHS becomes 0 since $x$ is fixed). – user29743 Apr 28 '12 at 16:59
For whatever reason I moved the $t$ to the left hand side, because at the time I was a little sketched out about differentiability. – user29743 Apr 28 '12 at 16:59
You can simply take the $\lim_{t \to 0}$ on the RHS and use FTC. – N. S. Apr 28 '12 at 17:12
@countinghaus: how does $f(x) = f(x+t)-f(t) \implies f(x) = cx$? – Kuku Apr 28 '12 at 18:02

$$f(x+y)= \frac{1}{t}\int_{0}^{t}[f(x+y+z)-f(z)]dz= \frac{1}{t}\int_{0}^{t}[f(x+y+z)-f(x+z)+f(x+z)-f(z)]dz= \frac{1}{t}\int_{0}^{t}[f(x+y+z)-f(x+z)]dz+\int_0^t[f(x+z)-f(z)]dz$$

for all $,x,y,t$.



After the substitution $u=x+z$ we get

$$f(x+y)-f(x)=\frac{1}{t}\int_{x}^{x+t}[f(y+u)-f(u)]du=\frac{1}{t}\left[\int_{0}^{x+t}[f(y+u)-f(u)]du -\int_{0}^{x}[f(y+u)-f(u)]du \right]=\frac{1}{t}\left[(x+t)f(y)-xf(y) \right]=f(y)$$



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This answer is meant to be a sort of stream-of-consciousness -- it's not meant to present the answer, but instead how you might follow your proof idea to obtain the answer. The thing we want to show is

$$ \begin{align} \frac{1}{t_1} \int_0^{t_1} f(x + y + z_1) - f(z_1) \, dz_1 &= \frac{1}{t_2} \int_0^{t_2} f(x + z_2) - f(z_2) \, dz_2 \\ &+ \frac{1}{t_3} \int_0^{t_3} f(y + z_3) - f(z_3) \, dz_3 \end{align}$$

for all $x, y$, and we are allowed to choose $t_1, t_2, t_3$ (and the choices may even depend on $x$ and $y$)

(to avoid confusing them, I've disambiguated the dummy variables! It turns out I didn't need to -- but at this point we don't yet know if we need the freedom!)

Well, to make the right side look like the left side, we could make a substitution. $z_2 = y + z_1$ will make $f(x + y + z_1)$ appear on the right hand side, so let's do that. (I could use $z_4$ to disambiguate some more, but I don't want it this time)

$$ \begin{align} \frac{1}{t_1} \int_0^{t_1} f(x + y + z_1) - f(z_1) \, dz_1 &= \frac{1}{t_2} \int_{-y}^{t_2 - y} f(x + y + z_1) - f(y + z_1) \, dz_1 \\ &+ \frac{1}{t_3} \int_0^{t_3} f(y + z_3) - f(z_3) \, dz_3 \end{align}$$

So that first integral on the right almost looks like the one on the left. Let's split it into two parts: one with the same integrand, and one for what's left over:

$$ \begin{align} \frac{1}{t_1} \int_0^{t_1} f(x + y + z_1) - f(z_1) \, dz_1 &= \frac{1}{t_2} \int_{-y}^{t_2 - y} f(x + y + z_1) - f(z_1) \, dz_1 \\ &- \frac{1}{t_2} \int_{-y}^{t_2 - y} f(y + z_1) - f(z_1) \, dz_1 \\ &+ \frac{1}{t_3} \int_0^{t_3} f(y + z_3) - f(z_3) \, dz_3 \end{align}$$

Let's choose $t_2 = t_1$ and cancel out what we can of those integrals:

$$ \begin{align} \frac{1}{t_1} \int_{t_1 - y}^{t_1} f(x + y + z_1) - f(z_1) \, dz_1 &= \frac{1}{t_1} \int_{-y}^{0} f(x + y + z_1) - f(z_1) \, dz_1 \\ &- \frac{1}{t_1} \int_{-y}^{t_1 - y} f(y + z_1) - f(z_1) \, dz_1 \\ &+ \frac{1}{t_3} \int_0^{t_3} f(y + z_3) - f(z_3) \, dz_3 \end{align}$$

Let's do the same with the last two integrals; substitute $z_3=z_1$ and $t_3=t_1$, and cancel:

$$ \begin{align} \frac{1}{t_1} \int_{t_1 - y}^{t_1} f(x + y + z_1) - f(z_1) \, dz_1 &= \frac{1}{t_1} \int_{-y}^{0} f(x + y + z_1) - f(z_1) \, dz_1 \\ &- \frac{1}{t_1} \int_{-y}^{0} f(y + z_1) - f(z_1) \, dz_1 \\ &+ \frac{1}{t_1} \int_{t_1-y}^{t_1} f(y + z_1) - f(z_1) \, dz_1 \end{align}$$

I'll now write $t_1=t$ and $z_1=z$. We can now simplify by multiplying through by $t$. With nothing else obvious to do, let's combine the integrals over the same ranges now. (I'm bringing the last one on the right over to the left side)

$$ \int_{t - y}^{t} f(x + y + z) - f(y+z) \, dz = \int_{-y}^{0} f(x + y + z) - f(y+z) \, dz$$

Note that everything we've done so far has been reversible. This is an important point, because we're working backwards -- I haven't been working towards the thing I'm trying to prove. Instead, I started with the thing I'm trying to prove and saw what I could derive from that!

Explicitly, given the choice of $t_2$ and $t_3$ I made above, this last equation is true if and only if the thing we are trying to prove is true.

Unfortunately, we can't choose $t=0$ to finish the proof. (do you see why?) Let's simplify again, by substituting $z = z_4 - y$:

$$ \int_{t}^{t+y} f(x + z) - f(z) \, dz_4 = \int_{0}^{y} f(x + z) - f(z) \, dz_4$$

At this point, I see N.S.'s proof, and he shows the last trick needed:

$$ \int_{0}^{t+y} f(x + z_4) - f(z_4) \, dz_4 = \int_{0}^{y} f(x + z_4) - f(z_4) \, dz_4 + \int_0^t f(x+z_4) - f(z_4) \, dz_4$$

$$ (t+y) f(x) = y f(x) + t f(x)$$

One might have discovered this by recognizing we know things about $\int_0^{\star}$, so we should try writing $\int_{t}^{t+y}$ in terms of such things.

This last statement is obviously true. Now, by following the derivation in reverse, we eventually obtain the thing we wanted to prove.

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