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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 $0\leq t_1<t_2<...<t_n\leq 1$ 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.

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    $\begingroup$ Can you draw a function that is equal to $\mathrm{sign}(a_n)$ at the points $t_n$ and else lies between $[-1,1]$? $\endgroup$
    – s.harp
    Nov 24 '20 at 19:01
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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 $\|T\|=\sum_{k=1}^n|a_k|$.

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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 $\|T\| = \|a\|_1$.

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