# Find a linearly independent set of bounded linear functionals $\{f_1, \dots, f_n\}$ such that $f_i(x) = \alpha_i$ where $\alpha \in \Bbb R^n$

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?

• 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. – Matsmir Dec 29 '19 at 16:43
• @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? – Oliver G Dec 29 '19 at 19:45
• 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. – Matsmir Dec 29 '19 at 20:54
• @Matsmir Can you elaborate more on this? I'm having trouble understanding your notation. – Oliver G Dec 29 '19 at 21:11
• I have made an extended answer below. – 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.

• What is $\pi_i$ defined as? – Oliver G Dec 29 '19 at 17:05
• the coordinate projections. – Matematleta Dec 29 '19 at 17:06
• How is $\alpha_i(x)$ defined? – Oliver G Dec 29 '19 at 20:21
• 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, $a_i(x)=\pi_i\circ T(x)$ – Matematleta Dec 29 '19 at 22:30