Given a continuous mapping $f$ find $\lim_{n\to\infty}\int_0^1 nf(x)e^{-nx}$

Question

Given a continuous mapping $f: \Bbb{R} \to \Bbb{R}$ find $$\lim_{n\to\infty}\int_0^1 nf(x)e^{-nx}$$ Now I'm clueless, however I've had to solve a similar question $$\lim_{n\to\infty}\int_0^{\frac1{\sqrt n}}nf(x)e^{-nx}$$ I've did this by noting that since $f$ is continuous it attains it max/min on the interval $[0,\frac1{\sqrt n}]$, if we let $f(a)$ be the minimum and $f(b)$ the maximum then $$\int_0^{\frac1{\sqrt n}}nf(a)e^{-nx}= f(a)(-e^{-\sqrt n}+1)\leq \int_0^{\frac1{\sqrt n}}nf(x)e^{-nx}\leq f(b)(-e^{-\sqrt n}+1)=\int_0^{\frac1{\sqrt n}}nf(b)e^{-nx}$$ So $$\lim_{n\to \infty}f(a)(-e^{-\sqrt n}+1)\leq \lim_{n\to\infty}\int_0^{\frac1{\sqrt n}}nf(x)e^{-nx}\leq\lim_{n\to\infty}f(b)(-e^{-\sqrt n}+1)$$ So $$f(a)\leq\int_0^{\frac1{\sqrt n}}nf(x)e^{-nx}\leq f(b)$$ Since $a,b\in[0,\frac1{\sqrt n}]$ and $f$ is continuous means that $f(a)=f(b)=f(0)$. However I can't do the same for the $$\lim_{n\to\infty}\int_0^1 nf(x)e^{-nx}$$

Answer 1

HINT:

$$\int_0^1 ne^{-nx}f(x)\,dx=\int_0^\infty e^{-x}\xi_{x\in[0,n]}(x)f(x/n)\,dx$$

where $\xi_{x\in[0,n]}(x)$ is the indicator function. Now apply the Dominated Convergence Theorem.

Answer 2

Hint (without Dominated Convergence Theorem). Note that by the continuity of $f$ at $0$, for $\epsilon>0$ there is $0<\delta\leq 1$ such that $|f(x)-f(0)|\leq\epsilon$ for $x\in [0,\delta]$. Hence $$\begin{align} \left|\int_0^1 ne^{-nx}f(x)dx-f(0)\right|&=\left|\int_0^1 ne^{-nx}(f(x)-f(0))dx-f(0)e^{-n}\right|\\ &\leq \int_0^{\delta} ne^{-nx}|f(x)-f(0)|dx+\int_{\delta}^1 ne^{-nx}|f(x)-f(0)|dx+|f(0)|e^{-n}\\ &\leq (1-e^{-n\delta})\epsilon+(e^{-n\delta}-e^{-n})(\max_{x\in [0,1]}|f(x)|+|f(0)|)+|f(0)|e^{-n}. \end{align}$$ Can you take it form here?

Answer 3

I thought it might be instructive to present another way forward that circumvents the use of the Dominated Convergence Theorem. To that end, we now proceed.

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First, we note that for any $\epsilon>0$ there exists a $\delta>0$ such that $|f(x)-f(0)|<\epsilon$ whenever $|x|<\delta$. For a given $\epsilon>0$, fix such a $0<\delta<1$.

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Next, we write

$$\begin{align} \int_0^1 ne^{-nx}f(x)\,dx&=\int_0^1 ne^{-nx}f(0)\,dx+\int_0^1 ne^{-nx}\left(f(x)-f(0)\right)\,dx\\\\ &=f(0)(1-e^{-n})+\int_0^1 ne^{-nx}\left(f(x)-f(0)\right)\,dx\\\\ &=f(0)(1-e^{-n})\\\\ &+\int_0^\delta ne^{-nx}\left(f(x)-f(0)\right)\,dx\\\\ &+\int_\delta^1 ne^{-nx}\left(f(x)-f(0)\right)\,dx\tag1 \end{align}$$

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We can estimate the second term on the right-hand side of $(1)$ as

$$\left|\int_0^\delta ne^{-nx}\left(f(x)-f(0)\right)\,dx\right|\le \epsilon (1-e^{-n})\tag2$$

while we can estimate the third term as

$$\left|\int_\delta^1 ne^{-nx}\left(f(x)-f(0)\right)\,dx\right|\le 2\max_{x\in [0,1]}(f(x))\,(e^{-n\delta}-1)\tag3$$

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Taking the limit as $n\to \infty$ in $(2)$ and $(3)$ we find that

$$\lim_{n\to \infty}\left|\int_0^1 ne^{-nx}(f(x)-f(0))\,dx\right|\le \epsilon$$

Since $\epsilon>0$ was arbitrary, we must have

$$\lim_{n\to \infty}\int_0^1 ne^{-nx}(f(x)-f(0))\,dx=0$$

whence we find the coveted limit

$$\lim_{n\to \infty}\int_0^1 ne^{-nx}f(x)\,dx=f(0)$$