Linear Transformation (Integration)

Question

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Although I am not extremely clear on why it is a mapping to $C[0,1]$, I know how to prove that $L$ is a linear operator.

I am mostly having trouble understanding how to read this linear transformation and find $L(e^x)$ and $L(x^2)$

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Answer Key:

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Given a $L(e^x)$ for example, I know that to take integration, it must be a function of $t$, but why can we just make it $e^t$?

Also $F$ is a function of $x$ which is confusing me. I assume it has nothing to do with $f$ and has to do with the upper bound?

What is an easier way of interpreting this linear transformation?

Answer 1

By the fundamental theorem of calculus, if $f$ is continuous on $[0,1]$, the primitive

$$ P_f(t) = \int_0^tf(x)dx $$

will also be continuous. This justifies that the function $L(f) = P_f$ is well defined, and in fact linear:

$$ L(\lambda f+ \mu g)(t) = \int_0^t(\lambda f+\mu g)(x)dx = \int_0^t[\lambda f(x) +\mu g(x)]dx = \\ = \lambda\int_0^tf(x)dx +\mu\int_0^tg(x)dx = \lambda L(f)(t) + \mu L(g)(t) = (\lambda L(f) + \mu L(g))(t) $$

and therefore $L(\lambda f+ \mu g) = \lambda L(f) + \mu L(g)$.

The statement on the exersice should be a bit more clear, there is an abuse of variable use (which is then fixed in the answer sheet when they switch from $e^x$ to $e^t$).

Intuitively, $L$ is nothing more than integrating continuous functions, and the linearlity of integration is something we are very much used to work with. This is just formalizing this notion in the context of linear algebra.

Answer 2

Why $L(f)=F$ is in $C[0,1]$? For $x\in[0,1)$ and small $h>0$, one can use Mean Value Theorem to deduce that \begin{align*} F(x+h)-F(x)=\int_{0}^{x+h}f(t)dt-\int_{0}^{x}f(t)dt=\int_{x}^{x+h}f(t)dt=hf(\xi_{h}), \end{align*} where $\xi_{h}\in[x,x+h]$, as $h\rightarrow 0^{+}$, we have $f(\xi_{h})\rightarrow f(x)$ and hence $F(x+h)-F(x)\rightarrow 0$.

Similar reasoning is valid for $h<0$. And one can do it for $x=1$.

Answer 3

The problem (I think) is an abuse of notation.

The map $L$ is defined by $L(f)(x) = \int_0^x f(t) dt$.

For each $x \in [0,1]$, the quantity $\int_0^x f(t) dt$ is a number. Hence $L(f)$ defines a real valued function defined on $[0,1]$.

It is straightforward to see that it is linear because the integral is linear.

To see that $L(f)$ is continuous, note that since $f$ is continuous it is bounded by some number $B$. Then, if $x<y$ we have $L(f)(y)-L(f)(x) = \int_x^y f(t)dt$ and so $|L(f)(y)-L(f)(x)| \le B|x-y|$. Hence it is continuous.

I would write $L(x \mapsto e^x)$ to make it clear that the argument is a function. In this case we have $L(x \mapsto e^x)(y) = \int_0^y e^t dt = e^y -1$. Hence $L(x \mapsto e^x)$ is the function $x \mapsto e^x -1$.