$f'\in \mathcal R([0,1])$ , then $\lim_{n \to \infty} \sum_{k=1}^n f(\frac kn) - n \int_{0 }^1 f(x)dx=\frac{f(1)-f(0)}2$?

Question

If $f:[0,1]\to \mathbb R$ is a differentiable function with continuous derivative then I can show that

$$ \lim_{n \to \infty} \left[ \sum_{k=1}^n f\!\left(\dfrac kn \right) - n \int_0^1 f(x)\,dx \right] = \frac{f(1)-f(0)}2. $$

My question is: does this fact remain true if we only assume that $f$ is differentiable (on $[0,1]$)? Or just that the derivative of $f$ is bounded on $[0,1]$ and Riemann integrable?

(I know that just the continuity of $f$ does not lead to the identity.)

Answer 1

We will use the following result:

If $~g:[0,1]\to\mathbb{R}~$ has a Riemann integrable derivative. Then $$ \frac{g(0)+g(1)}{2}-\int_0^1g(x)dx= \int_{0}^1\left(x-\frac{1}{2}\right)g'(x)dx. $$

Indeed, having a Riemann integrable derivative implies that $g$ is a Lipschitz function, so it is absolutely continuous, and the formula of integration by parts, which is valid in this case, shows that: \begin{align*} \int_{0}^1\left(x-\frac{1}{2}\right)g'(x)dx &=\left.\left(x-\frac{1}{2}\right)g(x)\right]_{x=0}^{x=1} -\int_0^1g(x)dx\\ &=\frac{g(1)+g(0)}{2}-\int_0^1g(x)dx \end{align*}

Now applying this to the functions $x\mapsto f\left(\frac{k+x}{n}\right)$ for $k=0,1,\ldots,n-1$ and adding the resulting equalities we obtain $$ \sum_{k=0}^n f\left(\frac{k}{n}\right)-\frac{f(1)+f(0)}{2}-n\int_0^1f(x)dx= \int_0^1\left(x-\frac{1}{2}\right)H_n(x)dx\tag{1} $$ where, $$ H_n(x)=\frac{1}{n}\sum_{k=0}^{n-1}f'\left(\frac{k+x}{n}\right) $$ Clearly for every $x$, $H_n(x)$ is a Riemann sum of the Riemann integrable function $f'$, hence $$ \forall\,x\in[0,1],\quad\lim_{n\to\infty}H_n(x)=\int_0^1f'(t)dt $$ Moreover, $\vert{H_n(x)}\vert\leq\sup_{[0,1]}\vert{f'}\vert$. So, taking the limit as $n$ tends to $+\infty$, and applying the Dominated Convergence Theorem, we obtain $$\lim_{n\to\infty}\left( \sum_{k=0}^nf\left(\frac{k}{n}\right)-\frac{f(1)+f(0)}{2}-n\int_0^1f(x)dx\right)= \left(\int_0^1f'(t)dt\right)\int_0^1\left(x-\frac{1}{2}\right)dx=0. $$ which is the desired conclusion.

Answer 2

A proof without measure theory: Note that if $f'\in R[a,b],$ then $\int_a^b f' = f(b)-f(a).$ Proof: Briefly, the MVT shows

$$f(b)-f(a) = \sum_{k=1}^{n} (f(x_k) - f(x_{k-1})) = \sum_{k=1}^{n} f'(c(k,n))\Delta x_k \to \int_a^b f'.$$

Now for $f\in R[0,1],$ define

$$S_n(f) = \sum_{k=1}^{n}f(k/n) - n\int_0^1f.$$

Lemma: If $f'\in R[0,1],$ then

$$\mid S_n(f) \mid \le \int_0^1|f'|.$$

Proof: We can write

$$S_n(f) = n\sum_{k=1}^{n}\int_{(k-1)/n}^{k/n}(f(k/n) - f(t))\,dt =n\sum_{k=1}^{n}\int_{(k-1)/n}^{k/n}\int_{t}^{k/n}f'(s)\,ds\,dt.$$

Now take absolute values, replacing $f'(s)$ by $|f'(s)|$ and $t$ by $(k-1)/n.$ We get

$$|S_n(f)| \le \sum_{k=1}^{n}\int_{(k-1)/n}^{k/n}|f'(s)|\,ds= \int_0^1|f'|.$$

For the main result, assume $f'\in R[0,1].$ We use two facts. First, as you said, the result holds if $f'$ is continuous. Second, any Riemann integrable function can be approximated in the "Riemann norm" by continuous functions. So let $\epsilon>0.$ Then there exists $g$ continuous on $[0,1]$ such that $\int_0^1|f'-g| < \epsilon.$ Define $G(x) = \int_0^x g.$ We know the result holds for $G.$ So

$$S_n(f) - (f(1)-f(0))/2 = S_n(f-G) + [S_n(G)- (G(1)-G(0))/2] + [(G(1)-G(0))-(f(1)-f(0)]/2.$$

The last summand above equals $(1/2)\int_0^1(g-f').$ Taking absolute values and using the above results, we get

$$\limsup_{n\to \infty} |S_n(f) - (f(1)-f(0))/2| \le \epsilon + 0 + \epsilon/2.$$

Since $\epsilon$ is arbitrary, we have the desired limit.