Riemann-Lebesgue lemma for periodic functions [duplicate]

Suppose $f$ is a periodic function of period $T$ and $g$ is integrable on $[a,b]$. How do I prove that $$\lim_{t\to\infty} \int_a^b f(tx)g(x) \, dx = \frac 1 T \int_0^T f(x) \, dx \int_a^b g(x)\,dx$$

The problem didn't indicate that $f$ is continuous.

marked as duplicate by José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 23 '18 at 23:26

• If $f$ is periodic of period $T$ then $\frac{1}{T}=f$ where $f$ is frequency (Think Fourier Series). Then $\frac{1}{T}\int_{0}^{T}f(x)\,dx$ is like the $a_{0}$ term. – JohnColtraneisJC Apr 6 '18 at 13:01
• If $t$ tends to infinity on the LHS, the angular frequency $\omega$ gets very large which means the period $T$ becomes smaller and smaller. This means that on the RHS of the equation $\frac{1}{T}$ will be a very large spike. Since the period is very short $f(x)$ gets dominated by $g(x)$. Now if I could just formalize this into a proof... – JohnColtraneisJC Apr 6 '18 at 13:16
• But since $T$ is very small we know $\int_{0}^{T}f(x)dx$ will tend to 0 which is the result we would expect from our lemma. – JohnColtraneisJC Apr 6 '18 at 13:41
• Does the question allow any further assumptions on $f$? Integrability perhaps, to make the RHS make sense? – B. Mehta Apr 7 '18 at 0:33
Suppose $f$ is a periodic function of period $T$ and $g$ is integrable on $[a,b]$. We will show that $$\lim_{t\to\infty} \int_a^b f(tx)g(x) \, dx = \frac 1 T \int_0^T f(x) \, dx \int_a^b g(x)\,dx.$$ Proof: We are given $f$ is integrable. Denote $f_{avg}=\displaystyle{\frac{1}{T}\int_0^Tf(x)\,dx}$ as the average of $f$ over $[0,T]$. Denote $F$ as the anti-derivative of $f$. Then we have \begin{align*} F(x) &= \displaystyle {\bigg(\frac{1}{T}\int_0^Tf(t)\,dt\bigg)}x+g(x),\\ & \text{where $T>0$ is the period of $f$ and $g:\mathbb{R}\to\mathbb{R}$ is a $T$-periodic function.} \\ &\text{Lemma (1)} \end{align*} To prove this lemma, we may note that for any periodic function $f$, $f(t+T)=f(t)$ holds true for any $t\in\mathbb{R}$, so it follows that $F(x+t)-F(x)=\int_0^T f(t)\,dt$ for all $x\in\mathbb{R}$. We consider the function $h(x)=\displaystyle{\bigg(\frac{1}{T}\int_0^T f(t)\,dt}\bigg)x,$ we have $h(x+T)-h(x)=\int_0^Tf(t)\,dt$; and with our previous result $F(x+T)-h(x+T)=F(x)-h(x).$ This is the function defined by $g(x)=F(x)-h(x),\,x\in\mathbb{R}$, and is periodic of period $T$. This confirms lemma (1). Now using our lemma, we define our function $F(x)=f_{avg}\cdot\,x+h(x)$ for $x\in[a,b]$ where $h$ is continuous and has period $T$. We will write \begin{align*} & t\int_0^T\,g(x)f(tx)\,dx = \int_0^T\,g(x)F'(tx)\,dx = g(x)F(tx)\bigg|_{0}^{T}-\int_0^T\,g'(x)F(tx)\,dx=\\ &=g(T)F(tT)-\int_0^T\,g'(x)F(tx)\,dx = g(T)(f_{avg}tT+h(tT))-\\&-tf_{avg}\int_0^T\,xg'(x)\,dx-\int_0^T\,g'(x)h(tx)\,dx = tf_{avg}Tg(T)-\\&-tf_{avg}Tg(T)+tf_{avg}Tg(T)+tf_{avg}\int_0^T\,g(x)\,dx-\int_0^T\,g'(x)h(tx)\,dx = \\ &= tf_{avg}\int_0^T\,g(x)\,dx-\int_0^T\,g'(x)h(tx)\,dx. \end{align*} That is equivalent to \begin{align*} t\bigg(\displaystyle{\int_0^T\,g(x)f(tx)\,dx-\frac{1}{T}\int_0^T\,f(x)\,dx\int_0^T\,g(x)\,dx\bigg)= -\int_0^T\,g'(x)h(tx)\,dx.} (2) \end{align*} Then by the Riemann-Lebesgue lemma we have \begin{align*} & \displaystyle{\lim_{t\to\infty}\int_0^T\,g'(x)h(tx)\,dx = \frac{1}{T}\int_0^Th(x)\,dx\int_0^T\,g'(x)\,dx} = \\ &= \frac{1}{T}(g(T)-g(0))\int_0^T\,h(x)\,dx\int_0^T\,g'(x)\,dx = \\ &=\frac{1}{T}(g(T)-g(0))\bigg(\int_{0}^{T}F(x)\,dx-F(T)\bigg) \end{align*} and using (2) we have our result. $\blacksquare$
• Could you clarify ? It seems you are using $g$ (and $f$) to denote several different functions, in your notation we have $F(x)=f_{avg}x+g=f_{avg}x+f=\int_0^x f$. Also, if $f$ and $g$ are only $L^1_{loc}$, what does mean $\int fg$ ? – user120527 Apr 13 '18 at 7:42
• @user120527 I edited the bounds of the integral defining the average of $f$, and I also edited the grammatical syntax and argumentation style in certain lines to try to clarify. Would you prefer if I used different notation for the average of $f$ instead of $f_{avg}$? – JohnColtraneisJC Apr 13 '18 at 12:29