Can anyone help me with this? I want to know how to solve it.

Let $f:\mathbb R \longrightarrow \mathbb R$ be a continuous function with period $P$. Also suppose that $$\frac{1}{P}\int_0^Pf(x)dx=N.$$ Show that $$\lim_{x\to 0^+}\frac 1x\int_0^x f\left(\frac{1}{t}\right)dt=N.$$

  • 1
    $\begingroup$ That isn't true. E.g., let $f(x)=\sin(x)+1$, which has $P=2\pi$, and $N=1$, but $0\leq f(x)\leq 2$ for all $x$ implies $0\leq\int_0^xf(1/t)dt\leq 2x$ for all $x$. $\endgroup$ – Jonas Meyer Feb 2 '12 at 3:12
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    $\begingroup$ @Jonas: Indeed any bounded $f$ with nonnegative $N$ is a counterexample. Perhaps OP forgot to multiply the integral by $1/x$? $\endgroup$ – anon Feb 2 '12 at 3:25
  • $\begingroup$ I believe that the restriction on $f$ can be lightened from continuous to locally integrable. $\endgroup$ – robjohn Feb 4 '12 at 15:25

Note that if $f\in L^1([0,P])$, then $$ F(x)=\int_0^x(f(t)-N)\;\mathrm{d}t\tag{1} $$ is a continuous function of period P.

Then, with $M=\max\limits_{[0,P]}|F|$, we have

$$ \begin{align} \lim_{x\to0^+}\frac 1x\int_0^x f\left(\frac{1}{t}\right)dt &=\lim_{x\to0^+}\frac1x\int_\frac1x^\infty\frac{f(t)}{t^2}\mathrm{d}t\tag{2a}\\ &=\lim_{x\to\infty}x\int_x^\infty\frac{f(t)}{t^2}\mathrm{d}t\tag{2b}\\ &=N+\lim_{x\to\infty}x\int_x^\infty\frac{f(t)-N}{t^2}\mathrm{d}t\tag{2c}\\ &=N+\lim_{x\to\infty}x\int_x^\infty\frac{1}{t^2}\mathrm{d}F(t)\tag{2d}\\ &=N+\lim_{x\to\infty}x\left(-\frac{F(x)}{x^2}+2\int_x^\infty\frac{F(t)}{t^3}\mathrm{d}t\right)\tag{2e}\\ &=N\tag{2f} \end{align} $$

$\hskip{5mm}(\mathrm{2a})$ change of variables $t\mapsto1/t$

$\hskip{5mm}(\mathrm{2b})$ change of variables $x\mapsto1/x$

$\hskip{5mm}(\mathrm{2c})$ $\displaystyle N=x\int_x^\infty\frac{N}{t^2}\mathrm{d}t$

$\hskip{5mm}(\mathrm{2d})$ $\mathrm{d}F(t)=(f(t)-N)\;\mathrm{d}t$ follows from $(1)$ and the Fundamental Theorem of Calculus

$\hskip{5mm}(\mathrm{2e})$ integration by parts

$\hskip{5mm}(\mathrm{2f})$ follows from the simple bound $\displaystyle\left|-\frac{F(x)}{x^2}+2\int_x^\infty\frac{F(t)}{t^3}\mathrm{d}t\right|\le\frac{2M}{x^2}$

  • $\begingroup$ Nice, and doesn't need the density trigonometric polynomials. $\endgroup$ – Davide Giraudo Feb 4 '12 at 15:31
  • $\begingroup$ I don't understand how you got the forth and fifth line in the series of equalities. would you please be a bit more clear? $\endgroup$ – John Tesh Feb 7 '12 at 19:33
  • $\begingroup$ @John: I have explained each step in $(2)$. $\endgroup$ – robjohn Feb 7 '12 at 20:03

Using the function $g(t)=f\left(\frac P{2\pi}t\right)$, we can assume that $f$ is $2\pi$-periodic. If $P_n$ is a sequence of trigonometric polynomial which converges uniformly to $f$, then $$\left|\frac 1x\int_0^xf\left(\frac 1t\right)dt-\frac 1{2\pi}\int_0^{2\pi}f(t)dt\right|\leq 2\sup_{s\in\mathbb R}|f(s)-P_n(s)|+\left|\frac 1x\int_0^xP_n\left(\frac 1t\right)dt-\frac 1{2\pi}\int_0^{2\pi}P_n(t)dt\right|,$$ so, by linearity, it's enough to show the result when $f$ is of the form $e^{ipt}$, where $p\in\mathbb Z$. It's clear if $p=0$, so we assume $p\neq 0$. We have $\int_0^{2\pi}f(t)dt=\frac 1p(e^{i2\pi p}-1)=0$ and $$\frac 1x\int_0^xf\left(\frac 1t\right)dt=\frac 1x\int_{1/x}^{+\infty}\frac{e^{ips}}{s^2}dx=\int_1^{+\infty}\frac{e^{ip\frac yx}}{y^2}dy,$$ so \begin{align*} \frac 1x\int_0^xf\left(\frac 1t\right)dt&=\left[\frac x{ip}\frac{e^{ip\frac yx}}{y^2}\right]_1^{+\infty}-\frac x{ip}\int_1^{+\infty}\frac{e^{ip\frac yx}}{y^3}(-2)dy\\ &=-\frac x{ip}e^{ip/x}+2\frac x{ip}\int_1^{+\infty}\frac{e^{ip\frac yx}}{y^3}dy, \end{align*} and finally $$\left|\frac 1x\int_0^xf\left(\frac 1t\right)dt\right|\leq \frac x{|p|}\left(1+2\int_1^{+\infty}t^{-3}dt\right)=2\frac x{|p|},$$ which gives the result.


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