# A trapezoidal approximation for a sequence of uniformly converging functions

Let $f_n : [0,1] \rightarrow \mathbb R$ be a sequence of continuous functions converging uniformly to a function $f$. For each $N \in \mathbb N$, also define a sequence of points $0=x_1< x_2 <...<x_N=1$.

Does it hold that:

$$\lim_{N \rightarrow\infty} \sum_{k=1}^{N} \frac{f_N(x_{k-1}) + f_N(x_k)}{2} (x_k - x_{k-1}) = \int_0^1 f(x) dx$$

I have an intuition that this should hold, but I have no idea how to prove it.

• You haven't said enough about this sequence $x_k$ – zhw. Dec 11 '17 at 23:42
• My bad. I think a uniform partition would suffice. – Wolfups Dec 11 '17 at 23:58

Since $f_n(x) \to f(x)$ uniformly on $[0,1]$ with $f_n$ continuous, it follows that $f$ is integrable. As long as the partition $P_N = (x_0,x_1, \ldots, x_N)$ is constructed for each $N$ in such a way that the norm $\|P_N\| \to 0$ as $N \to \infty$, then any tagged Riemann sum $S(P_N,f,T)$ converges to the integral regardless of the choice of tags.

In particular, this convergence is true for left- and right-hand sums as well as the average of the two: $$\tag{1}\lim_{N \to \infty}A(P_N,f) = \lim_{N \to \infty} \sum_{k=1}^N \frac{1}{2}(f(x_{k-1}) + f(x_{k}))(x_k - x_{k-1}) = \int_0^1 f(x) \, dx.$$

Note that

$$\tag{2}A(P_N,f_N) = \sum_{k=1}^N \frac{1}{2}(f_N(x_{k-1}) + f_N(x_{k}))(x_k - x_{k-1}) \\= \sum_{k=1}^N \frac{1}{2}\{[f_N(x_{k-1})-f(x_{k-1})] + [f_N(x_{k})-f(x_k)]\}\,(x_k - x_{k-1}) + \sum_{k=1}^N \frac{1}{2}(f(x_{k-1}) + f(x_{k}))(x_k - x_{k-1}).$$

Hence, using (2) and applying the triangle inequality we have

$$\left|A(P_N,f_N) - \int_0^1 f(x) \, dx\right| \\ \leqslant \frac{1}{2}\sum_{k=1}^N |f_N(x_{k-1})-f(x_{k-1})|\, (x_k - x_{k-1}) + \frac{1}{2}\sum_{k=1}^N |f_N(x_{k})-f(x_k)|\,(x_k - x_{k-1}) +\left|A(P_N,f) - \int_0^1 f(x) \, dx\right|.$$

By uniform convergence, for any $\epsilon > 0$ there exists $N_1$ such that if $N > N_1$, then $|f_N(x) - f(x)| < \epsilon/2$ for every $x \in [0,1]$ and

$$\frac{1}{2}\sum_{k=1}^N |f_N(x_{k-1})-f(x_{k-1})|\, (x_k - x_{k-1}) + \frac{1}{2}\sum_{k=1}^N |f_N(x_{k})-f(x_k)|\,(x_k - x_{k-1}) < \epsilon/2,$$

since $\sum_{k=1}^N(x_k - x_{k-1}) = 1$.

By the convergence in (1) there exists $N_2$ such that if $N > N_2$, then $$\left|A(P_N,f) - \int_0^1f(x) \, dx\right| < \epsilon/2.$$

Therefore, if $N > \max(N_1,N_2)$ we have $|A(P_N,f_N) - \int_0^1 f(x)\, dx | < \epsilon,$ implying that

$$\lim_{N \to \infty}A(P_N, f_N) = \int_0^1 f(x) \, dx.$$

• @Wolfups: Your conjecture is true whether or not the partitions are uniform as long as the sequence of partition norms $\|P_n\| = \max_{1 \leqslant k \leqslant n} |x_k - x_{x-1}|$converges to $0$. – RRL Dec 12 '17 at 2:33

Hint: Define, for any $g$ on $[0,1],$

$$S_N(g) = \sum_{k=1}^{N}\frac{g((k-1)/N) + g(k/N)}{2}\frac{1}{N.}$$

Then $S_N(g)| \le \|g\|_\infty.$ Furthermore, if $g$ is Riemann integrable on $[0,1],$ then $S_N(g) \to \int_0^1 g.$

So consider in your problem $S_N(f_N) - S_N(f) = S_N(f_N-f).$