# numerical integration bounded linear operator

$$Let~S,~T_n~:(C[0,1],~||~||_∞)→(R,~|~|)~be~a~linear~operator.$$

$$S :=\int_{0}^{1}f(x) dx \\$$

$$T_n := \frac{1}{n}(\frac{1}{2}f(0) + \sum_{k=1}^{n-1} f(\frac{n}{k}) + \frac{1}{2}f(1))$$

$$Show~that~S,~T_n~are~bounded~operators~and~||S-T_n||~=~2$$

I solved S and T_n are bounded, but I have no idea how to solve the latter.

We can write $$T_nf=\int f d\mu$$ where $$\mu =\frac 1 {2n} \delta_0+\frac 1 {2n} \delta_1+ \sum\limits_{k=1}^{n-1}\frac 1 n \delta_{n/k}$$. Let $$m$$ be Lebesgue measure on $$[0,1]$$ and $$\nu =m-\mu$$. Then $$(S-T_n)(f)=\int f d\nu$$. By Riesz Representation Theorem the norm of $$S-T_n$$ is nothing but the total variation of the measure $$\mu=m-\mu$$. Since $$m \perp \mu$$ it follows that $$\|S-T_n\|=\|m\|+\|\mu\|=1+ \sum\limits_{k=1}^{n-1}\frac 1 n+\frac 1 {2n}+\frac 1 {2n}=2$$.

Hints for constructing an elementary proof: $$\|T_n-S\|\leq \|T_n\|+\|S\| \le 1+1=2$$. Given $$\epsilon >0$$ construct a piece-wise linear continuous function $$f$$ of norm $$1$$ such that $$f(x)=-1$$ for $$x \in\{0,1,\frac 1n, \frac 2 n,..., \frac{n-1} n\}$$ and $$\int f(x) dx >1-\epsilon$$. Then $$\|T_n-S\|\geq |Sf-T_nf| >2-\epsilon$$.

• @e osa I have added another proof. Jul 8, 2020 at 7:13
• Thanks. I totally understand. Jul 8, 2020 at 11:35
• Could you tell me why ∫𝑓(𝑥)𝑑𝑥>1−𝜖 ? For such a function, ∫𝑓(𝑥)𝑑𝑥 = 1, I think. Jul 8, 2020 at 12:17
• You cannot just define $f$ to be $-1$ at a finite number of points and $1$ at all other points. This is becasue $f$ has to be continuous. You have to use an approximation by a continuous function and that is why you cannot get $\int f(x)dx=1$. @eosa Jul 8, 2020 at 12:23

I think there are complicated ways to get to a solution, but I want to give you an easy straight-forward one:

First, you should know that the triangle inequality also holds for linear bounded operators. Therefore, we have $$|| S-T_n || \leq ||S|| + ||T_n|| = 1+1=2 .$$

Second, as you surely know, the next step normally involves finding a non-zero function $$f$$ in $$C([0,1],\mathbb{R})$$ such that $$|(S-T_n)f|=2 \cdot ||f||.$$

Unfortunately, this isn't possible.

But there is another way to get the result, namely to find a sequence of non-zero functions $$f_m$$ such that $$\frac {| (S-T_n)f_m |} {||f_m||} \xrightarrow {k \to \infty} 2.$$

Now you start by considering the case $$n=1$$. Just start by sketching the functions you'll need. The solution in this case is a sequence of functions that have the values $$f(0)=-1=f(1)$$ and immediately go to $$y=1$$ with increasing steepness.

PS: To make it easier, consider functions with norm $$1$$.