# Let $V = C[0, 1]$ and define an operator $T: V \rightarrow V$ by the following formulas for $T(f)$ as a function of the variable $x$.

I am reading the book, Applied Linear Algebra and Matrix Analysis.
When I was doing the exercise of Section3.1 Exercise 13, I was puzzled at some answers of it. Here is the problem description:

Let $$V = C[0, 1]$$ and define an operator $$T: V \rightarrow V$$ by the following formulas for $$T(f)$$ as a function of the variable $$x$$. Which of these operators is linear?
If so, is the target V of the operator equal to its range?
$$(a)f(1)x^2$$ $$(b)f^2(x)$$ $$(c)2f(x)$$ $$(d)\int^x_0f(s)ds$$

I think (a)$$f(1)x^2$$ is not linear just because:

$$g(u) = f(1)u^2,\ g(v) = f(1)v^2 \\ g(cu) = f(1)c^2u^2 = c^2f(1)u^2 \\ cg(u) = cf(1)u^2 \neq g(cu) \\$$
The range is $$V$$ and here is my derivation:

$$\because V = C[0, 1] \\ \therefore x, f(x) \in C[0,1] \\ \therefore f(1) \in C[0,1] \\ \therefore f(1)x^2 \in C[0,1]$$

But the answer of it is:

$$(a)$$linear,range not $$V$$ $$(b)$$ not linear, $$(c)$$ linear, range is $$V (d)$$ linear, range not $$V$$

So the answer of (a) is not like what I thought before.
I think the range of $$(c)2f(x)$$ is not $$V$$.
Also I think the range of $$(d)\int^x_0f(s)ds$$ is $$V$$ because the integral of $$f(s)$$ on the range$$[0,x]$$ is a area which is less than or equal to a area consisting of (0, 0) and (1, 1).
If not mind, could anyone tell my thoughts is right or not and give me some advice on how to determine it is the linear operator and its range on such questions?

• I suspect you mean $T:V\to V$? – blub Apr 1 at 12:15

Linear means $$T(af+bg)(x)=aTf(x)+bTg(x)$$. (You don't change $$x$$ to $$cx$$; you only change $$f$$ to $$cf$$ etc). It should now be clear that b) is not linear and the other three are linear. The range in a) and d) are functions with some special properties: in a) they vanish at $$0$$ and in d) they are differentiable. So the range is not $$V$$ in these cases. It is clear that the range is $$V$$ in c).

• Dear Sir, I still have some questions after reviewing some relevant knowledge. If you don't mind, could you describe more specific on The range in a) and d) are functions with some special properties: in a) they vanish at 0 and in d) they are differentiable. So the range is not V in these cases. and give a direction about it. – Bowen Peng Apr 1 at 12:53
• If $g$ is a function which does not vanish at $0$ then it cannot be written as $T(f)$ for any $f$. So range of $T$ is a proper subset of $C[0,1]$. – Kavi Rama Murthy Apr 1 at 13:11

The operator is linear if and only if $$T(f+g)$$ is the same function as $$T(f)+T(g)$$. To determine whether this is true, you need to calculate, for a pair of function $$f,g$$, what the function $$T(h+g)$$ is, and what $$T(f)+T(g)$$ is.

The function $$T(f)$$ maps $$x$$ to $$f(1)\cdot x^2$$, and the function $$T(g)$$ maps $$u$$ to $$g(1)\cdot x^2$$.

The function $$T(f+g)$$, therefore, maps $$u$$ to $$(f+g)(1)\cdot x^2$$.

Can you continue from here? Remember:

1. in $$V$$, the definition of addition is such that $$(f+g)(x)=f(x)+g(x)$$, which should help you calculate $$(f+g)(1)$$.
2. From that, you can write, in terms of $$f$$ and $$g$$, where $$T(f+g)$$ maps $$x$$.
3. Then compare that to where $$T(f)+T(g)$$ maps $$x$$ (use rule from point 1 again)