Why is this function almost Lipschitz?

Question

We are still in the saga of solving the 2002 qualifier. This question 6b has stumped me and I am mostly clueless about it:

Say $f:\mathbb{R}\rightarrow \mathbb{R}$ is bounded with a finite constant $B$ such that: $$\frac{|f(x+y)+f(x-y)-2f(x)|}{|y|}\leq B$$ Prove there exists $M(\lVert f \rVert_\infty, B)$ such that for all $x\not=y$: $$|f(x)-f(y)|\leq M |x-y|\left(1+\ln_+(\frac{1}{|x-y|})\right)$$ Where $\ln_+(x)=\max \{0,\ln(x)\}$

Intuitively this means that away from $y=x$, $f(x+y)\rightarrow f(x)$ linearly. Close to $x$, it is still true $f(x+y)\rightarrow f(x)$ but it is slightly perturbed by $\ln_+$. Here are a couple of facts which have gotten me nowhere:

Fact 0. The inequality in $B$ would be an approximation for $f''(x)$ if it were divided by $y^2$ instead of $y$.

This is particularly useless, because we have no regularity associated with $f$. Even if we did $|f''(x)|\leq \lim M/|y|=\infty$ so this observation cannot be of any help.

Fact 1. $\lim_{y\rightarrow 0} |f(x-y)-f(x)|$ exists.

One has by inverted triangle inequality that:

$$\lim_{y\rightarrow 0}||f(x+y)-f(x)|-|f(x-y)-f(x)||\leq \lim_{y\rightarrow 0 }B|y|=0$$

But taking away the modulus and changing $\lim$ to $\limsup$: $$\lim_{y\rightarrow 0}\sup_{y\in[-a,a]}(|f(x+y)-f(x)|-|f(x-y)-f(x)|)=0$$

$$\lim_{a\rightarrow 0}\sup_{y\in [-a,a]}|f(x+y)-f(x)|\leq\lim_{a\rightarrow 0}\inf_{y\in [-a,a]}|f(x-y)-f(x)|$$

But as we are taking the infimum in a symetric interval, hence we may write: $$\lim_{a\rightarrow 0}\sup_{y\in [-a,a]}|f(x+y)-f(x)|\leq\lim_{a\rightarrow 0}\inf_{y\in [-a,a]}|f(x+y)-f(x)|\leq \lim_{a\rightarrow 0}\sup_{y\in [-a,a]}|f(x+y)-f(x)|$$

$$\lim_{a\rightarrow 0}\sup_{y\in [-a,a]}|f(x+y)-f(x)|=\lim_{a\rightarrow 0}\inf_{y\in [-a,a]}|f(x+y)-f(x)|$$

This means that $\lim |f(x+y)-f(x)|$ exists.

Fact 2. There is the obvious $y=y-x$ substitution

$$\frac{|f(y)+f(2x-y)-2f(x)|}{|y-x|}\leq B$$ $$|f(y)-f(x)|\leq B|y-x|+|f(2x-y)-f(x)|$$

Answer 1

Set $K = \Vert f \Vert_\infty$. In order to simplify the notation a bit we fix $y \in \Bbb R$ and consider the function $$ g: \Bbb R \to \Bbb R , \, g(u) = f(y+u) - f(y) \, . $$ Note that $\Vert g \Vert_\infty \le 2K$. We will prove that there is a constant $M = M(K, B)$ such that $$ \tag{$*$} |g(u)| \le M |u| \left(1+ \log^+ \frac {1}{|u|} \right) $$ for all $u \ne 0$. Then $$ |f(x) - f(y)| = |g(x-y)| \le M |x-y| \left(1+ \log^+ \frac {1}{|x-y|} \right) $$ for all $x \ne y$.

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Proof of $(*)$: For $|u| \ge 1$ is $$ |g(u)| \le 2K \le 2K|u| \, , $$ that is the desired linear upper bound $(*)$ with $M=2K$.

Now we consider the case $0 < |u| < 1$. We have $$ \begin{align} 2 |g(u)| - |g(2u)| &\le |g(2u) - 2g(u)| \\ &= | f(y+2u) + f(y) - 2 f(y+u)| \\ &\le B |u| \, , \end{align} $$ using the given inequality with $\tilde x = y+u$ and $\tilde y = u$. It follows that $$ |g(u)| \le \frac 12 |g(2u)| + \frac{B|u|}{2} \, . $$ We can apply this repeatedly to $2u, 4u, 8u, \ldots$: $$ \begin{align} |g(u)| &\le \frac 12 \left( \frac 12 |g(4u)| + B|u| \right) + \frac{B|u|}{2} = \frac 14 |g(4u)| + B|u| \\ &\le \frac 14 \left( \frac 12 |g(8u)| + 2B|u| \right) + B|u| = \frac 18 |g(8u)| + \frac{3B|u|}{2} \\ &\le \cdots \\ &\le \frac 1{2^n} |g(2^nu)| + \frac{nB|u|}{2} \le \frac {2K}{2^n} + \frac{nB|u|}{2} \end{align} $$ for all nonnegative integers $n$. Setting $n = \lfloor \log(1/|u|)\rfloor$ gives $$ |g(u)| \le 4 K |u| + \frac{B|u|}{2} \log \frac{1}{|u|} = |u| \left( 4K + \frac B2 \log \frac{1}{|u|} \right) $$ and that is $(1)$ with $M=\max(4K, B/2)$.

Combining the results for $0 < |u| < 1$ and $|u| \ge 1$ we see that $(*)$ holds for all $ u \ne 0$ with $M=\max(4K, B/2)$, and that concludes the proof.

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Remarks:

• The above proof is inspired by proofs of “midpoint convex + bounded function $\implies$ continuous”, e.g. here or here.

• As in the case of midpoint convex functions, the boundedness condition is crucial (compare this). Any non-continuous solution of Cauchy's functional equation $f(x+y) = f(x) + f(y)$ satisfies the given condition (with $B=0$), but not the conclusion. Such non-continuous solutions exists, and they are necessarily unbounded.