# Norm of $T:(C([0,1]),||\cdot||_\infty)\rightarrow\mathbb{R}$ where $Tf=\sum_{k=1}^n a_kf(t_k)$?

I need to calculate the operator Norm of the linear operator defined as: $$T:(C([0,1]),||\cdot||_\infty)\rightarrow\mathbb{R} \text{ where } Tf=\sum_{k=1}^n a_kf(t_k)$$ for $$0\leq t_1 and $$a_1,...,a_n\in \mathbb{R}$$.

I have been able to show that $$||T||\geq\left|\sum_{k=1}^n a_k\right|$$ and $$||T||\leq \sum_{k=1}^n|a_k|$$ but I don't seem to able to bound it further than that. Would appreciate any help. Thank you.

• Can you draw a function that is equal to $\mathrm{sign}(a_n)$ at the points $t_n$ and else lies between $[-1,1]$? Nov 24 '20 at 19:01

WLOG, one can assume that $$a_k\ne0$$ for all $$k$$ (if not, just drop all the zero-values, and in the case that all are zero, the problem is trivial).

Take $$f(x)$$ a continuous function with these properties:

• $$f(t_k)=\frac{|a_k|}{a_k}$$.
• $$|f(x)|\le 1$$ for all $$x\in[0,1]$$.

For example, you can take the polygonal through the points $$\{(t_k,f(t_k)),\,1\le k\le n\}\cup\{(s_k,0),1\le k\le n-1\}$$, where $$s_k=\frac{t_k+t_{k+1}}{2}$$ (the mid point of the interval $$[t_k,t_{k+1}]$$).

Then it is clear that $$\|f\|_\infty=1$$ (observe that $$|f(t_k)|=1$$) and $$Tf=\sum_{k=1}^n |a_k|$$.

This together with your calculations leads to $$\|T\|=\sum_{k=1}^n|a_k|$$.

It might help to separate $$T$$ into the composition of two simpler maps.

Note that the map $$s:C[0,1] \to \mathbb{R}^n$$ defined by $$s(f) = (f(t_1),...,f(t_n))$$ has norm one (using the $$l_\infty$$ norm on $$\mathbb{R}^n$$) and is surjective. Furthermore, for any $$y$$ with $$\|y\|_\infty =1$$ it is easy to construct (by interpolation & extrapolation) some $$f \in C[0,1]$$ with $$\|f\| = 1$$ such that $$s(f) = y$$.

Let the operator $$\tau:\mathbb{R}^n \to \mathbb{R}$$ be given by $$\tau(y) = \sum_k a_k y_k$$. It is straightforward to show that $$\|\tau\| = \|a\|_1$$ and there is some $$y \in \mathbb{R}^n$$ such that $$\|y\|_\infty = 1$$ and $$\tau(y) = \|a\|_1$$.

Note that $$T = \tau \circ s$$.

Hence $$\|T\| \le \|s\| \|\tau\| = \|a\|_1$$. Suppose $$\tau(y) = \|a\|_1$$ with $$\|y\|_\infty = 1$$ and $$s(f) = y$$ where $$\|f\| = 1$$, then $$Tf = \tau(s(f)) = \tau(y) = \|a\|_1$$ and so $$\|T\| = \|a\|_1$$.