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Let $f:[0,1]\longrightarrow \mathbb{R}$ be a continuous function. Prove, that $$f(0) = \lim_{x\longrightarrow 0^{+}} x \int_x^1 \frac{f(t)}{t^2}\ dt$$ It may have to do with fundamental theorem of calculus, but I'm not able to grasp the connection yet.

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But then how do you show it is $f(0)$?? –  tattwamasi amrutam Jun 14 at 11:42

3 Answers 3

up vote 3 down vote accepted

Let $\epsilon>0$.

Choose $0<\delta<1$ so that $ f(x)<f(0)+\epsilon$ for all $0\le x<\delta$.

Write $$ x\int_x^1{f(t)\over t^2}\,dt = \underbrace{x\int_x^\delta{f(t)\over t^2}\,dt}_{A(x)}+ \underbrace{x\int_\delta^1{f(t)\over t^2}\,dt}_{B(x)}. $$

Since $f(x)/x^2$ is bounded on $[\delta,1]$ we have $\lim\limits_{x\rightarrow 0^+} B(x)=0$.

Also, for $0<x<\delta$, $$\eqalign{ A(x) &= x\int_x^\delta {f(t)\over t^2}\,dt\cr &\le x\int_x^\delta {f(0)+\epsilon\over t^2}\,dt\cr &=x(f(0)+\epsilon)\Bigl( {1\over x}-{1\over\delta}\Bigr)\cr &=(f(0)+\epsilon)\Bigl(1-{x\over\delta}\Bigr) . } $$ It follows that $$\tag{1}\limsup\limits_{x\rightarrow0^+} x\int_x^1{f(t)\over t^2}\,dt\le f(0)+\epsilon.$$

In a similar manner, one can show $$\tag{2} \liminf\limits_{x\rightarrow0^+} x\int_x^1{f(t)\over t^2}\,dt\ge f(0)-\epsilon. $$

Since $\epsilon$ was arbitrary, it follows from $(1)$ and $(2)$ that $$\lim\limits_{x\rightarrow0^+} x\int_x^1{f(t)\over t^2}\,dt=f(0).$$

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Using a change of variables, it suffices to prove the equivalent statement

$$f(0)=\lim_{x \to 0^+} x \int_{1}^{1/x} f(1/u)du .$$

Step 1: The statement clearly holds for all monomials $f(x)=1,x,x^2,\dots$.

Step 2: Since the problem is linear, it holds for all polynomial $f(x)$.

Step 3: Use the Stone-Weierstraß Theorem: Let $\varepsilon>0$ be given, and find a polynomial $p$ which approximates $f$ uniformly in the interval $[0,1]$ up to $\varepsilon$, and agrees with $f$ at $0$. We then have

$$x \int_1^{1/x}f(1/u)du=x \int_1^{1/x} (f(1/u)-p(1/u))du+x \int_{1}^{1/x}p(1/u)du $$

The last term gives us $p(0)=f(0)$ in the limit, so all we need it to show that the first term tends to $0$ as $x \to 0^+$. Indeed:

$\left\vert x \int_1^{1/x} (f(1/u)-p(1/u))du \right\vert \leq x \int_1^{1/x} \left\vert f(1/u)-p(1/u) \right\vert du \leq x \varepsilon(1/x-1)$ which tends to $\varepsilon$ as $x \to 0^+$.

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Let $$F(x)=\int_x^1\frac{f(t)}{t^2} \, dt.$$ Since $f$ is continuous on $[0,1]$ hence it has a minimum value on this interval. Call it $m$. Then $$\int_x^1\frac{f(t)}{t^2} \, dt \geq m \int_x^1\frac{1}{t^2} \, dt=m\left(\frac{1}{x}-1\right).$$ But as $x \to 0^{+}$, the last expression approaches $\infty$.

Now you can apply L'Hospital to the limit $$\lim_{x\longrightarrow 0^{+}} \frac{\int_x^1 \frac{f(t)}{t^2}\ dt}{1/x}$$ Also note that by the FTC $$\frac{d}{dx}\int_x^1 \frac{f(t)}{t^2}\ dt=-\frac{f(x)}{x^2}$$

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what if the number $m \leq 0$? –  Paramanand Singh Jun 14 at 13:29

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