Is the limit of the mean value of a function around a point equal to the value of the function at that point?

Question

Suppose that the function $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous. Is the following statement correct? $$\lim_{n \rightarrow \infty}\frac{n}{2} \int_{y-\frac{1}{n}}^{y+\frac{1}{n}}f(x) \text{d}x=f(y)$$ If it is, how to prove it? If it isn't, what would be a counterexample?

Answer 1

Yes, that is correct. By the integral mean value theorem, $$ \int\limits_{y-1/n}^{y+1/n}f(x)\,dx=2f(c)/n $$ where $c\in (y-1/n,y+1/n)$. As $n\to\infty$, $c\to y$ and, by continuity, $f(c)\to f(y)$.

Answer 2

It's true since \begin{align} |\frac{1}{2\delta}\int_{y - \delta}^{y + \delta}f(x)\,dx - f(y)| &= |\frac{1}{2\delta}\int_{y - \delta}^{y + \delta}f(x) - f(y)\,dx| \\ &\leq \frac{1}{2\delta}\int_{y - \delta}^{y + \delta}|f(x) - f(y)|\,dx \\ &\leq \frac{1}{2\delta}\sup_{x \in (y - \delta, y + \delta)}|f(x) - f(y)| \cdot 2\delta \\ &= \sup_{x \in (y - \delta, y + \delta)}|f(x) - f(y)| \to 0 \text{ as }\delta \to 0. \end{align} We know the last quantity goes to $0$ since that is the definition of continuity at $y$.

Answer 3

Let's rewrite in terms of $h = \frac1{n}.$ As $n \to \infty, h \to 0^+,$ so we get

$$\lim_{n \to \infty} \frac{n}2 \int_{y - 1/n}^{y + 1/n} f(x) dx = \lim_{h \to 0^+} \frac{1}{2h} \int_{y - h}^{y + h} f(x) dx$$

Now by basic integral properties we can rewrite this as

$$\lim_{h \to 0^+} \frac{\int_0^{y + h} f(x) dx - \int_0^{y - h} f(x) dx}{2h} = \frac{d}{ds}\left[\int_0^s f(x) dx\right]_{s = y}$$

where the equivalency of the alternative definition of the derivative can be justified either by L'Hospital's rule or with limit operations. Now, because $f$ is continuous, by the Fundamental Theorem of calculus we have that $\frac{d}{ds}\left[\int_0^s f(x) dx\right]_{s = y} = f(y).$