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Let $f:\mathbb R\to\mathbb R$ be continuous such that $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}~\forall~x,y\in\mathbb R.$ How to show that $f$ is a straight line?

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The two points $(x,f(x))$ and $(y,f(y))$ are on the graph of $f$. The point on the straight line half-way between them is $\left(\frac{x+y} 2, \frac{f(x)+f(y)} 2 \right)$. The actual point on the graph with abscissa $\frac{x+y}2$ is $\left(\frac{x+y} 2, f\left(\frac{x+y} 2 \right) \right)$. And we are told those two points are the same.

Now iterate this: We’ve shown the conclusion that’s it’s a straight line is right for $x$ and $y$ and the point $z$ half-way between them: those three points on the graph are on a straight line. Now deal in the same way with the points half-way between $x$ and $z$ (call it $a$), and half-way between $z$ and $y$ (call it $b$). So far we have

$$

x < a < z < b< y.

$$

Then half-way between $x$ and $a$; then half-way between $a$ and $z$; then half-way between $z$ and $b$; then half-way between $b$ and $y$.

Then further split the resulting intervals in half. Then split the next set of resulting intervals in half. And so on.

No matter how long we continue the process, the points we getting by splitting smaller and smaller intervals in half will thus be shown always to lie on a straight line.

However, some points between $x$ and $y$ will never be reached by successively splitting in half: the point two-thirds of the way from $x$ to $y$ is one such point. Here we use continuity of the function $f$: that point can be approached as closely as you wish by points that do result from successively splitting intervals in half.

Thus the part of the graph between $(x,f(x))$ and $(y,f(y))$ is a straight line.

But that is true no matter which two points are chosen to be called $x$ and $y$. Hence the whole graph is a straight line.

Like this answer it can be easily proved that $$f(ax+by) =af(x) + bf(y)\tag{1}$$ for all real $x, y$ and all positive real $a, b$ with $a+b=1$. It is easily seen that any function which satisfies $(1)$ has the following property: if $f$ vanishes at two distinct points then it vanishes at every point between these two points. Now let $k$ be a positive and fixed real number and let $$g(x) = \frac{f(k) – f(0)}{k}\cdot x + f(0)$$ Then clearly $g$ is linear and satisfies the functional equation $(1)$. It is also seen that if two functions satisfy $(1)$ then their sum as well as their difference also satisfies $(1)$. It follows that the function $h(x) =f(x) – g(x)$ satisfies $(1)$ and clearly $h(k) =h(0)=0$ therefore $h(x) =0$ for all $x\in[0,k]$. Thus $$f(x) =g(x) = \frac{f(k) – f(0)}{k}\cdot x +f(0)$$ for all $x\in[0,k]$. Since $k$ was arbitrary it follows that $f$ is a linear function on whole of $\mathbb{R} $.

Here is another simpler approach. Let $g(x) = f(x) – f(0)$ then we can easily check that $$g\left(\frac{x + y}{2}\right) = \frac{g(x) + g(y)}{2}$$ and we have $g(0) = 0$. Now we can see that $g(x) = g((2x + 0)/2) = (g(2x) + g(0))/2$ or $g(2x) = 2g(x)$ for all $x$. Next we can see that $$g(x + y) = g\left(\frac{2x + 2y}{2}\right) = \frac{g(2x) + g(2y)}{2} = g(x) + g(y)$$ This is the standard Cauchy’s functional equation and since $g$ is continuous we can see that $g(x) = xg(1)$ and therefore $$f(x) = g(x) + f(0) = xg(1) + f(0) = x(f(1) – f(0)) + f(0)$$ so that $f$ is a linear function.

Define a linear function $g \colon \mathbb{R} \rightarrow \mathbb{R}$ by $g(x) = (f(1) – f(0))x + f(0)$. Note that by definition, $g(0) = f(0)$ and $g(1) = f(1)$ and that $g$ also satisfies

$$ g \left( \frac{x + y}{2} \right) = \frac{g(x) + g(y)}{2}. $$

You want to show that $g(x) = f(x)$ for all $x \in \mathbb{R}$. We have

$$ f \left( \frac{x + 0}{2} \right) = \frac{f(x) + f(0)}{2}. $$

Plugging in $x = 1$, we see that

$$ f \left( \frac{1}{2} \right) = \frac{f(1) + f(0)}{2} = \frac{g(1) + g(0)}{2} = g \left( \frac{1}{2} \right) $$

(where we used $f(0) = g(0)$ and $f(1) = g(1)$). Plugging in $x = \frac{1}{2}$ will give you

$$ f \left( \frac{1}{4} \right) = \frac{f \left( \frac{1}{2} \right) + f(0)}{2} = \frac{g \left( \frac{1}{2} \right) + g(0)}{2} = g \left( \frac{1}{4} \right)$$

(where we used $f(0) = g(0)$ and $f \left( \frac{1}{2} \right) = g \left( \frac{1}{2} \right)$). Iterating this inductively, you can show that

$$ f \left( \frac{1}{2^n} \right) = g \left( \frac{1}{2^n} \right) $$

for all $n \in \mathbb{N}$. By a similar inductive argument, you can then show that

$$ f \left( \frac{m}{2^n} \right) = g \left( \frac{m}{2^n} \right) $$

for all $n \in \mathbb{N}$ and $m \in \mathbb{Z}$. Since $f$ and $g$ are two continuous functions that agree on the dense set $ \{ \frac{m}{2^n} \, | \, n \in \mathbb{N}, m \in \mathbb{Z} \}$ of the dyadic rationals, they must agree everywhere and so $f(x) = g(x)$ for all $x \in \mathbb{R}$.

Set

$$f(\frac{t}{2})=\frac{f(t-x)}{2}+\frac{f(x)}{2}$$

This is valid for any $x$ and any $ t$. So, we have

$$2f(\frac{t}{2})=f(t-x)+f(x)$$

Express $f(t)$ as $f(t)=ktg(t)+c$ for some function $g(t)$. This makes it

$$2k\frac{t}{2}g(\frac{t}{2})=k(t-x)g(t-x)+kxg(x)$$

This brings us to

$$tg(\frac{t}{2})=tg(t-x)+x(g(x)-g(t-x))$$

We take now $x=0$ since this above has to be valid for any $x$ and this is what we have in the end.

$$g(\frac{t}{2})=g(t)$$

If there are two points $a,b$ for which $$g(a)\neq g(b)$$ then $$g(\frac{a}{2^n})\neq g(\frac{b}{2^n})$$ and this would mean that $g$ is not continuous at $0$. As $n$ grows large we can get as close as we want to $0$.

Because $g(t)$ is continuous, it is constant and from there $f(x)$ must have a linear form $f(x)=bx+c$.

A Jensen function is both Jensen-convex and Jensen-concave. It could be easily checked (by dyadic convex combinations) that a continuous Jensen-convex (Jensen-concave) function is convex (concave). Hence our $f$ is both convex and concave, which means that $f\bigl(tx+(1-t)y\bigr)=tf(x)+(1-t)f(y)$ (for any $x,y\in\mathbb{R}$ and $t\in[0,1]$). This implies that the graph of $f$ is a straight line.

I agree; that ‘some points between x and y will never be reached by successively splitting in half: the point two-thirds of the way from x to y is one such point’ if by that you mean directly.

But Jensens equation can express that claim indirectly ie F(2/3x)=2/3F(x) with F(0)=0,

and F in-jective and (strict perhaps) increasing where F:[0,1]to [0,1] for example.

I might be wrong, but it can clear express F(3x)=3F(x) and consequently F(1/3x)=1/3F(x), cant it? Unless that step is an invalid substitution (just take the midpoint o

f (A) F(4x)=4F(x) and (B) F(2x)=2F(x),o get that (C)F(3x)=3F(x); and thus (C.1) F(1/3x)=1/3F(x) through careful substitution.

Which is akin to saying that the F(1/3F(1))=1/3F(1) lies a third of the way between F(0)=min=0 and F(1)=max and indirectly as a result F(2/3F(1))=2/3F(1) using F(1/2x)=1/2F(x) That is before any kind of significant continuity or regularity assumptions are required.

Perhaps you mean if F is not strictly monotonic for example and its not designated F(0)=0 and F:[0,1]to [0,1]

(A) follows with F(0)=0 and Jensens equation), and (B), which is a consequence of (A). Using the jensens equation on these points to get (c) and (c1). And then one use use F(1/3x)=1/3F(x) and F(2x)=2F(x) to express F(2/3x)=F(2\times (1/3x)) using (A) =2F(1/3\timesx) and then by(C1). F(2/3x)= =2\times1/3F(x)=2/3F(x) and (C1). Likewise this is some sense akin to expresses a difference,albeit indirect, between the minimum element F(0)=0 and F(1)=max.element in the co-domain.

I apologize if I am miscontrue-ing you, as this is important,and so I need to be told about my error if so. And I find the post quite interesting for other reasons

. I agree that it cannot really directly express; and not in the same kind of way that cauchy’s equation can make these claims about rationals. But it appears that it can still capture them, before one requires a stronger regularity requirement to capture the irrational values.

That is before any kind of significant continuity or regularity assumptions are required. Perhaps you mean if F is not strictly monotonic for example and its not designated F(0)=0 and F:[0,1]to [0,1].

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