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, 2019 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, 2019 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, 2019 at 20:54
  • $\begingroup$ @Matsmir Can you elaborate more on this? I'm having trouble understanding your notation. $\endgroup$
    – Oliver G
    Dec 29, 2019 at 21:11
  • $\begingroup$ I have made an extended answer below. $\endgroup$
    – Matsmir
    Dec 29, 2019 at 21:59

2 Answers 2


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, 2019 at 17:05
  • $\begingroup$ the coordinate projections. $\endgroup$ Dec 29, 2019 at 17:06
  • $\begingroup$ How is $\alpha_i(x)$ defined? $\endgroup$
    – Oliver G
    Dec 29, 2019 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, $a_i(x)=\pi_i\circ T(x)$ $\endgroup$ Dec 29, 2019 at 22:30

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