given that $h$ is differentiable and $h(3x)+h(2x)+h(x) = 0$ for all $x\in\mathbb{R},$ is $h$ necessarily identically zero?

Question

Let $h:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that $h(3x)+h(2x)+h(x) = 0$ for all $x\in\mathbb{R}.$ Determine with proof whether $h\equiv 0.$

I think $h$ is identically zero.

Let $(1)$ be the original equation. Suppose for a contradiction that there exists some $x$ for which $h(x)\neq 0.$ Then differentiating (1) gives $3 h'(3x) + 2h'(2x)+h'(x)=0$ for all $x\in \mathbb{R}.$ Clearly $h(0) = 0,$ which can be seen by substituting $x=0$ into $(1)$.

Claim 1: For any $\epsilon > 0,$ and any $t\neq 0, |\dfrac{h(t)}{t}| < \epsilon.$

I'm not sure how to prove claim 1, but if it's true, then it immediately follows that h is identically zero.

Answer 1

This is basically a modified version of Ryszard's solution, but maybe a little simpler.

Since $h(x)$ is differentiable, $g(x) = h(x)/x$ with $g(0)=h'(0)=0$ is continuous.

Dividing $h(x)+h(2x)+h(3x)=0$ by $x$ gives $$ g(x)+2g(2x)+3g(3x)=0. $$ This makes $|3g(3x)|=|g(x)+2g(2x)|$, which means either $|g(x)|\ge|g(3x)|$ or $|g(2x)|\ge|g(3x)|$.

If we have some $t_0$ for which $g(t_0)\not=0$, using the above result for $3x=t_0$ allows us to pick $t_1$ equal to either $t_0/3$ or $2t_0/3$ so that $|g(t_1)|\ge|g(t_0)|$. Repeat this step, and you get a sequence $t_0,t_1,\ldots\rightarrow0$ so that $0<|g(t_0)|\le|g(t_1)|\le\cdots$ which contradicts $g(t_i)\rightarrow g(0)=0$ as $g$ is continuous.

Answer 2

$$h(3x) + h(2x) + h(x) = 0.$$

If $h(x) \rightarrow \ell$ as $x \rightarrow \infty$ then $\ell = 0$. Hence $|h(y)| < \epsilon$ for all $y > B$ for some $B$. Now $$h(3y) + h(2y) + h(y) = 0.$$ Assuming $|h(3y)| < \epsilon, |h(2y)| < \epsilon$ => $|h(y)| < 2 \epsilon$. Combining above conditions: $|h(B)| < 2 \epsilon$. Hence $|h(y)| < 3 \epsilon$ for all $y \geq B/2$. Let $|h(y)| < 2^t \epsilon$ for all $y \geq B/2^t $. This now implies, $|h(y)| < 2^{t+1} \epsilon$ for all $y > B/2^{t+1}$. Now use the continuity at $0$ and $h(0) = 0$ and do a similar argument as above and get $|h(y)| < 2^{t+1} \epsilon$ for all $y \leq (\frac{3}{2})^{t+1} \delta$. Now $\frac{B}{2^{t^*+1}} = (\frac{3}{2})^{t^*+1} \delta$ =>$B = 3^{t^*+1} \delta$ => $t^* = \log_3(B/\delta)-1$. Hence $|h(y)| < 2^{t^*+1} \epsilon$ for all $y$. Note that $B,\delta$ depends on $\epsilon$.

Now using only continuity at $0$ since continuity at $\infty$ has to be proved: $$|h(y)| < 2^{t+1} \epsilon \text{ for all } y \leq \left(\frac{3}{2} \right)^{t+1} \delta$$ $$|h(y)| < \frac{3^{t+1} \delta}{y} \epsilon \text{ for all } y$$ $$\frac{|h(y)|}{y} < \frac{3^{t+1} \delta}{y^2} \epsilon \text{ for all } y$$

Substituting $t+1 = \log_{1.5}(3) \log_3(\frac{y}{\delta})$ in above inequality, $$\frac{|h(y)|}{y} < \frac{(\frac{y}{\delta})^{\log_{1.5}(3)-1}}{y} \epsilon \text{ for all } y$$ $$\frac{|h(y)|}{y} < \frac{y^{\log_{1.5}(3)-2}}{\delta^{\log_{1.5}(3)-1}} \epsilon \text{ for all } y$$

Hence $|h(y)|$ is atmost cubic in $y$, using only continuity at $0$ and the functional equation.

Answer 3

It's false. Take $h(x)=\Re( e^{z\ln x})$ for a constant $z\in\mathbb C$.

$h(3x)+h(2x)+h(x)=\Re\left [(e^{z\ln 3} +e^{z\ln 2} + 1)e^{z\ln x}\right]$

We only need to find a $z\in\mathbb C$ such that $$e^{z\ln 3} +e^{z\ln 2} + 1=0$$

There exists such a $z$ and WolframAlpha says that $z\simeq -0.454+3.598i$.

Edit: As mentioned in comments, I need this defined on the whole line. So maybe we can look at $h(x)=\Re( e^{z\ln |x|})$, but I can't guarantee that it is differentiable at $x=0$.

Answer 4

The condition implies $h(0)=0$ and by differentiability $h'(0)=0.$ The continuity at the points $x\neq 0$ is not necessary for the conclusion to hold.

Fix $\varepsilon >0.$ There exists $\delta>0$ such that $|h(x)|\le \varepsilon |x| $ for $|x|\le \delta.$ Let $$M(x)=\sup_{0<|t|\le x}{|h(t)|\over t},\quad x>0$$ Along the proof it will be shown that $M(x)$ is finite. Then $0\le M(x)\le \varepsilon$ for $ |x|\le \delta.$ By assumptions for $0<|t|\le x$ we have $${h(t)\over t}= -{{2\over 3}}\,{h({{2\over 3}}t)\over {2\over 3}t}-{1\over 3}{h({1\over 3}t)\over {1\over 3}t}$$ Therefore $$M(x)=\sup_{0<|t|\le x}{|h(t)|\over t}\le {2\over 3}M({\textstyle {2\over 3}}x)+{1\over 3}M({\textstyle{1\over 3}x})\qquad (*)$$ which means $M(x)$ is less or equal to a convex combination of $M( {2\over 3}x)$ and $M( {1\over 3}x).$ There exists $N$ such that ${2^N\over 3^N}x\le \delta.$ By iterating $N$ times inequality $(*)$ we obtain $$M(x)\le \sum_{k=1}^K\lambda_kM(x_k),\quad \lambda_k>0,\ \sum_{k=1}^K\lambda_k=1,\ 0<|x_k|\le \delta$$ for some points $\{x_k\}_{k=1}^K.$ In particular $M(x)$ is finite and $$M(x)\le \varepsilon$$ As $\varepsilon>0$ is arbitrary, we may conclude that $M(x)=0$ for $x>0,$ hence $h(x)\equiv 0.$

Remark It suffices that $h$ is differentiable at $0.$ The solution is valid for $\beta \ge 1,$ $\alpha\ge \beta+1$ and $$h(\alpha x)+h(\beta x)+h(x)=0 \quad (*) $$ For the case $\alpha<\beta+1$ the conclusion holds under additional assumptions: $\beta^2+1\le \alpha^2,$ $h$ differentiable, and $h$ differentiable twice at $0.$ Indeed differentiating $(*)$ gives $$\alpha h'(\alpha x)+\beta h'(\beta x)+h'(x)=0$$ Then $${|h'(t)|\over t}\le {\beta^2\over \alpha^2}{|h'({\textstyle {\beta\over \alpha}t})|\over {\beta\over \alpha}t}+{1\over \alpha^2}{|h'({\textstyle {1\over \alpha}t})|\over {1\over \alpha}t}$$ Now we can apply the reasoning in the answer with $h'$ in place of $h,$ provided that $\beta^2+1\le \alpha^2.$ In this way we get $h'(x)\equiv 0,$ i.e. $h$ is constant, thus $h\equiv 0.$ For example the conclusion holds for $\alpha={5\over 2}$ and $\beta =2.$

Answer 5

Here's yet another one.

Clearly $h(0)=0$. Now for every $t\in{\mathbb R}$,

\begin{align*} (*)\quad \frac{h(t)}{t} = \frac{2}{3}(-\frac{h(\frac{2t}{3})}{\frac{2t}{3}}) + \frac{1}{3}(-\frac{h(\frac{t}{3})}{\frac{t}{3}})\end{align*}

We're going to use it to prove two things:

1. By taking the limit $t\to0$, it follows that

$$h'(0)= - h'(0)~\Rightarrow ~h'(0)=0.$$

2. Fix any $t$. As the righthand side of $(*)$ is a convex combination, we conclude that

$$ |\frac{h(t)}{t}|\le |\frac{h(c t)}{ct}|,$$

for some $c\in (0,\frac23)$. Since $(\frac{2}{3})^n \to 0$, by iterating this we have that for every $n\in{\mathbb N}$, there exists some $c_n \in (0,(\frac{2}{3})^n)$ so that

$$ |\frac{h(t)}{t}|\le |\frac{h'(c_n t)}{c_n t}|.$$

Taking $n\to\infty$, the RHS tends to $|h'(0)|$, which is equal to $0$ by part 1.