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Let $S$ be a set, $F$ a field, and $V(S;F)$ the space of all functions from $S$ into $F$: $$ (f + g)(x) = f(x) +g(x)$$ $$(cf)(x) = cf(x)$$ Let $W$ be any $n$-dimensional subspace of $V(S,F)$ . Show that there exist points $x_1, x_2, ... x_n$ in $S$ and functions $f_1, f_2, ..., f_n$ in $W$ such that $f_i(x_j) = \delta_{ij}$.

This question has been asked before and the proof of the problem involved induction. However I think this can be solved without using induction.

Attempt :

Define $T_x : V(S;F)\to F$ as $T_x(f)=f(x)$ where $x\in S.$ It is trivial to see that $T_x$ is a linear transformation. Now consider the restriction of $T_x$ on $W.$

Let $\{f_1,\dots,f_n\}$ be a basis for $W.$ Using an elementary theorem we know there exists linear transformations $L_i : W\to F$ such that $L_i(f_j)=\delta_{ij}$ for each $i, 1\le i\le n.$

If I am able to show $L_i=T_{x_i}$ for some $x_i\in S$ then I am done.

I need help completing the proof this way OR please point a flaw showing why the proof cannot be going this way.

Bijesh K.S
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  • why would you avoid induction? it is a nice mathematical tool – supinf Jul 20 '17 at 11:32
  • @supinf This problem is mentioned in the section "The Double dual," so there must be a way to do this involving it. I was just trying to do it that way. – Bijesh K.S Jul 21 '17 at 08:07

2 Answers2

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It is not always possible to show $L_i=T_{x_i}$ for some $x_i$.

Consider $S=\mathbb R = F$. Choose $W=\{f : f(1)=f(0), f(x)=0 \forall x\in \mathbb R\setminus\{0,1\} \}$ which is a $1$-dimensional subspace. A basis of $W$ is given by $f_1(x)=\delta_{x,1} + \delta_{x,0}$.

$L_i$ can be defined as $\frac12 (T_1+T_0)$. Thus it cannot be $T_{x_i}$ for some $i\in\mathbb R$.

supinf
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  • Your answer is incorrect because $L_1$ can instead be defined as $T_0$. When $W$ is finite-dimensional we can always find $x_i$ in $S$ such that $L_i = T_{x_i}$. –  Jan 13 '18 at 16:57
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From the comments above.

Since the proof is constructive, induction is a natural option.

However, if you wish you can avoid the language of induction by modifying the way you write the proof. See this question in which this is done.

Moreover, the other answer by @supinf is incorrect, because $L_1$ can instead by defined as $T_0$. When $W$ is finite-dimensional we can always find $x_i$ in $S$ such that $L_i = T_{x_i}$.