From Functional Analysis by Kreyszig:

If a linear operator $T:X \rightarrow Y$ on a normed space has a finite dimensional range. show that $T$ has a representation of the form $$Tx = \sum_{i=1}^n f_i(x)y_i$$ where $\{y_1, \dots, y_n\}$ and $\{f_1, \dots, f_n\}$ are linear independent sets in $Y$ and the dual space $X'$.

I can see that since $T(X)$ is finite, $T(X)$ has basis $B = \{y_1 = T(x_1), \dots, y_n = T(x_n)\}$ and therefore $T(x) = \sum_{i=1}^n\alpha_iy_i$, where $y_i = T(x_i)$ where $\{x_i\}$ is a linearly independent set in $X$.

I know I have to find a linearly independent set of bounded linear functionals $\{f_i\}$ such that $f_i(x) = \alpha_i$, but I'm having trouble showing this.

Anyone have any ideas?

  • $\begingroup$ You need a set of functionals that satisfy $f_i(x_i) = \alpha_i$ and $f_i(x_j) = 0$ if $i \ne j$. Your functionals are well defined and (obviously) continuous on a finite-dimensional subspace. And now all you need is Hahn-Banach theorem. $\endgroup$ – Matsmir Dec 29 '19 at 16:43
  • $\begingroup$ @Matsmir If you assume that $f_i(x) = \sum_j \alpha_j \delta_{ij}$ then the Hahn-Banach theorem says there's an extension of $f_i : \text{span}\{x_1, \dots, x_n\} \rightarrow Y$ to $X \rightarrow Y$ such that $T(x) = \sum f_i(x) T(x_i)$ for $x$ in the span. But why does this guarantee that for $x$ outside of $\text{span}\{x_1, \dots, x_n\}$ that this relationship holds? $\endgroup$ – Oliver G Dec 29 '19 at 19:45
  • $\begingroup$ Yep, that's a mistake. Take $g_i$ - linear functionals on Y - such that $g_i(y_i)=\alpha_i$ and $g_i(y_j) =0$ for $i \ne j$ (make them in fashion of my previous comment). After that let $f_i(x) = g_i(Tx)$. I think now it works. $\endgroup$ – Matsmir Dec 29 '19 at 20:54
  • $\begingroup$ @Matsmir Can you elaborate more on this? I'm having trouble understanding your notation. $\endgroup$ – Oliver G Dec 29 '19 at 21:11
  • $\begingroup$ I have made an extended answer below. $\endgroup$ – Matsmir Dec 29 '19 at 21:59

Let's start from a general result. Assume that we have $L \subset Y$ where $Y$ is a normed space and $L$ is a finite-dimensional subspace. Let $y_1, \dots, y_n$ be a basis in $L$. Then there exist continuous linear functionals on $Y$ $g_1, \dots, g_n \in Y^*$ s.t. $g_i(y_j) = \delta_{ij}$. Also in this case operator $P:Y \rightarrow Y$ that is defined by $Py = \sum\limits_{i = 1}^n g_i(y)y_i$ satisfies following properties:

i) P is continuous

ii) range of $P$ is $L$

iii) $Py = y$ for all $y \in L$

I will prove the existence of functionals $g_i$. Properties of $P$ follow immediately. Existence of $g_i$ is a corollary of Hahn-Banach theorem: $g_i$ are defined on $L$ by equations $g_i(y_j) = \delta_{ij}$ and then they are extended to $Y$.

Next consider $T:X \rightarrow Y$ - continuous operator with finite-dimensional range $L = T(X) \subset Y$. Then apply previous results to $L$ and obtain linear functionals $g_1, \dots,g_n$ and $P:Y \rightarrow Y$ s.t. $Py = \sum\limits_{i = 1}^n g_i(y)y_i$, $y_i \in L$, $Py = y$ for all $y \in L$. Then you have $$ Tx = PTx = \sum\limits_{i = 1}^n g_i(Tx) y_i = \sum\limits_{i = 1}^n f_i(x) y_i $$ where $f_i = g_i \circ T$. $y_i$ are linearly independant by defenition, linear independance of $f_i$ is easy to check: you can find $x_i \in X$ s.t. $Tx_i = y_i$. Then $f_i(x_j) = \delta_{ij}$. This implies linear independance.


There are $(x_i)\subseteq X$ such that $\{T(x_i)\}^n_{i=1}$ is a basis for $T(X)$. Then, $T(x)=\sum^n_{i=1}a_i(x)T(x_i)$ and since $T(x_j)$ has a unique expression in this basis, we must have $a_i(x_j)=\delta^j_i$. Now, if there are scalars $(c_i)$ such that $c_1a_1+\cdots c_na_n=0$, then evaluating at $x_j$, we get $c_j\cdot 1=0\Rightarrow c_j=0$ so the $a_j$ are linearly independent. To finish, apply the coordinate projections to $T$. Then, $(a_i)=(\pi_i\circ T)$ are the desired functionals.

  • $\begingroup$ What is $\pi_i$ defined as? $\endgroup$ – Oliver G Dec 29 '19 at 17:05
  • $\begingroup$ the coordinate projections. $\endgroup$ – Matematleta Dec 29 '19 at 17:06
  • $\begingroup$ How is $\alpha_i(x)$ defined? $\endgroup$ – Oliver G Dec 29 '19 at 20:21
  • $\begingroup$ For each $x\in X$ there are scalars $(a_i)$ such that $T(x)=\sum^n_{k=1}a_iT(x_i)$. As $x$ varies, so do the $a_i$. Hence, there is a function $x\mapsto a_i(x)$ from $X$ to the scalar field. To be more precise, $a_i(x)=\pi_i\circ T(x)$ $\endgroup$ – Matematleta Dec 29 '19 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.