Suppose that f is a continuous real function on [0, 1]. Prove that
\[ \operatorname* { l i m } _ { \alpha \to 0 ^ { + } } \alpha \int _ { 0 } ^ { 1 } x ^ { \alpha - 1 } f ( x ) d x = f ( 0 ) . \]
Solution
Set \[ g(x)\coloneqq f(x)-f(0). \]
1. For every \(\alpha>0\),\[ \alpha\int_0^1 x^{\alpha-1}f(0)\,dx=f(0). \]
Since \(\alpha>0\), \[ \alpha\int_0^1x^{\alpha-1}\,dx = \alpha\left[\frac{x^\alpha}{\alpha}\right]_0^1 = 1. \] Multiplying by \(f(0)\) gives the claim.
2. One has\[ \lim_{\alpha\to0^+} \alpha\int_0^1x^{\alpha-1}g(x)\,dx = 0. \]
Let \(\varepsilon>0\). By continuity of \(g\) at \(0\) and \(g(0)=0\), choose \(\delta\in(0,1)\) such that \[ \abs{g(x)}<\frac{\varepsilon}{2} \] for \(0\leq x\leq\delta\). Since \(g\) is continuous on \([0,1]\), set \[ M\coloneqq\max_{0\leq x\leq1}\abs{g(x)}. \] Then, for every \(\alpha>0\), \[ \begin{aligned} \abs{ \alpha\int_0^1x^{\alpha-1}g(x)\,dx } &\leq \alpha\int_0^\delta x^{\alpha-1}\abs{g(x)}\,dx + \alpha\int_\delta^1x^{\alpha-1}\abs{g(x)}\,dx\\ &\leq \frac{\varepsilon}{2}\delta^\alpha + M(1-\delta^\alpha)\\ &\leq \frac{\varepsilon}{2} + M(1-\delta^\alpha). \end{aligned} \] Because \(0<\delta<1\), \[ \delta^\alpha\longrightarrow1 \] as \(\alpha\to0^+\). Hence, for all sufficiently small \(\alpha>0\), \[ M(1-\delta^\alpha)<\frac{\varepsilon}{2}. \] The displayed estimate is then \(<\varepsilon\), proving the limit.
3. Therefore\[ \lim_{\alpha\to0^+} \alpha\int_0^1x^{\alpha-1}f(x)\,dx = \boxed{f(0)}. \]
Since \(f=g+f(0)\), split the integral into these two terms. Step 1 evaluates the constant term, and step 2 shows that the remaining term tends to \(0\).
4. Q.E.D.
Step 3 is the required limit.