# Operator on a set spaned the space

1. May 19, 2012

### LikeMath

Hi there,

Let $X$ be a Hilbert (Banach) space, and spanned by a set $S$, say.
Let $A$ be linear bounded operator on X into itself.
Suppose that the operator is well known on S, that is
$Aa_i=b_i$ for all $a_i\in S$.
First, is this operator unique on X? if yes, can we find $Aa$, for general element a in X, in terms of $b_i$.

2. May 19, 2012

### micromass

Staff Emeritus
What do you mean with this?? I ask because you might mean something completely different that I am thinking right now.

3. May 19, 2012

### LikeMath

That is, the space is generated by the closure of all linear combinations from S.

4. May 19, 2012

### micromass

Staff Emeritus
Ah, I suspected something like that.

The answer is yes, the operator will be unique.

You can see that is two stages: suppose that T is unique on S, then it's also unique on linear combinations of S. Indeed, take a linear combination $a=\sum \lambda_i a_i$, then we have

$$T(a)=\sum \lambda_i T(a_i)$$

Then use that if T is unique on a dense subset, then it's unique on the entire set. Indeed, take x in the closure, then there is a sequence $x_n\rightarrow x$. It must hold that

$$T(x_n)\rightarrow T(x)$$

5. May 19, 2012

### LikeMath

Thank you.
I completely agree with you. But what about the second part of the question, that is
can we find the operator on the whole space X?