Prove $f\left(nx\right)\leq nf\left(x\right).$

Question

Probelm,

$f:\left [ 0,+\infty \right ) \to \mathbb{R}$ is continuous on $\left [ 0,1 \right ]$ and differentiable on $\left ( 0,1 \right ) $, and $f(x) = f(x+1),\; f(0) = 0$, $f^{\prime}(x)$ is monotonically decreasing on $\left ( 0,1 \right ) $, prove$f\left(nx\right)\leq nf\left(x\right)\; \forall x \in \left [ 0,+\infty \right )$ and $\forall n \in \mathbb{N}^{*}$

My attempts,

According to a property of Convex function,

If $f$ is a convex function of one real variable, and $f(0) \le 0$ then $f$ is superadditive on the positive reals, that is $f(a+b)\geq f(a)+f(b)$ for positive real numbers $a$ and $b$.

In this problem, $f$ is a concave function ,then $f(a+b)\le f(a)+f(b)$ when $a,b \in \left [ 0,1 \right ]$. \begin{aligned} \Rightarrow &\quad nf(x) \ge f(nx),\;\forall x \in \left [ 0,1 \right ] \\ \Rightarrow&\quad nf(x) \ge f(nx),\;\forall x \in \left [ 0,+\infty \right ) \end{aligned} The second step uses $f(x) = f(x+1)$.

But there is a mistake in my proof, after $x$ is fixed, when $n$ is sufficiently large, $nx$ will be greater than 1, and then I wondered if there was a counterexample, but I couldn't find it.

Answer 1

Fix two positive real number $ x $ and $ r $ and let $ F(a,b) = f(b) - f(a) $ for each non-negative real number $ a $ and $ b $, it's sufficient to show that $ F(r, r+x ) \leq f(x) $ (since one could put $ r = nx $).

Notice the following two properties:

(1). Suppose $ \delta < a < b $ such that $ (a-\delta, b) \subseteq [m,m+1] $ for some natural number $ m $, since $ f' $ is monotonically decreasing on $ (m, m+1 ) $, thus we have $ f'(t-\delta) \leq f'(t) $ for each $ t \in (a,b) $ and therefore \begin{aligned} F(a,b) &= \int_{a}^b f'(t) \mathrm{d}t \leq \int_{a}^b f'(t-\delta) \mathrm{d}t \\ &= \int_{a-\delta}^{b-\delta} f'(t) \mathrm{d}t = F(a-\delta,b-\delta). \end{aligned}

(2). $ F(m, a) = F(n, a) $ for each natural two numbers $ m $ and $ n $.

Therefore, if $ (r,r+x) \subseteq [m,m+1] $ for some natural number $ m $, then we have $$ F(r, r+x ) \leq F(r-(r-m), r+x-(r-m)) = f(x). $$

Otherwise, there are two natural numbers $ k, l $ such that $$ k \leq r \leq k+1 \leq l \leq r+x \leq l+1 .$$ Put $ \alpha = k+1-r $ and $ \beta = r+x-l $.

(i). If $ \alpha + \beta \leq 1 $, put $ \delta = 1 - \alpha - \beta $, then we have \begin{aligned} F(r, r+x ) =& F(k+1,l+\beta) + F(l+1-\alpha, l+1 ) \\ \leq& F(k+1,l+\beta) + F(l+1-\alpha-\delta, l+1-\delta ) \\ =& F(k+1, k+1+x ) = f(x). \end{aligned}

(ii). If $ \alpha + \beta > 1 $, put $ \delta = 1 - \alpha $, then we have \begin{aligned} F(r, r+x ) =& F(l+2-\alpha, l+1+\beta ) \\ \leq& F(l+2-\alpha-\delta, l+1+\beta-\delta) \\ =& F(k+1, k+1+x ) = f(x). \end{aligned}

Q.E.D.

Answer 2

It is convenient to extend $f$ to all of $\mathbb{R}$ by $1$-periodicity, so that $f(x) = f(x+1)$ holds for all $x \in \mathbb{R}$, and we do so.

Now we turn to proving the inequality. In doing so, note that we may assume $x \in (0, 1)$ by the $1$-periodicity of $f$ and the condition $f(0) = 0$. So, let $n \geq 1$ and $ x \in (0, 1)$. Also, define $a$ and $b$ by

$$ a = x - \frac{\lfloor nx \rfloor}{n}, \qquad b = \frac{1}{n} - a. $$

Then $f(nx) = f(na) = f(-nb)$. Moreover, since $f$ is concave on $[0, 1]$ and $a \leq x \leq 1-b$, it turns out that

$$ \min\{ f(a), f(-b) \} = \min\{ f(a), f(1-b) \} \leq f(x). $$

In light of this, it suffices to prove:

$$ f(na) \leq n f(a) \qquad \text{and}\qquad f(-nb) \leq nf(-b). $$

However, each of these inequalities easily follow from OP's observation.