Proving that ##T## is skew-symmetric, inner product is an integration.

[Hall]

Homework Statement

Let $C(0.1)$ be the real linear space of all real functions continuous on $[0,1]$ with inner product
$$
\langle f , \rangle = \int_{0}^{1} f(t) ~g(t) dt $$
Let $V$ be the subspace of all $f$ such that
$$
\int_{0}^{1} f(t) dt = 0$$

Let $T: V \to C(0,1)$ be the integration operator defined by
$$
T (f(x)) = \int_{0}^{x} f(t) dt$$

Prove that $T$ is skew-symmetric.

Relevant Equations

$T$ will be skew-symmetric if
$$
\langle T(f), g \rangle = - \langle f, T(g)\rangle$$

$\langle T(f), g \rangle = \int_{0}^{1} \int_{0}^{x} f(t) dt ~ g(t) dt$
As $\int_{0}^{x} f(t) dt$ will be a function in $x$, therefore a constant w.r.t. $dt$, we have
$\langle T(f), g \rangle = \int_{0}^{x} f(t) dt ~ \int_{0}^{1} g(t) dt$

$\langle f, T(g)\rangle = \int_{0}^{1} f(t) ~ \int_{0}^{x} g(t) dt~ dt$
By the same argument, we have
$\langle f, T(g)\rangle = \int_{0}^{1} f(t) dt \int_{0}^{x} g(t) dt$

(Can someone tell me why preview button no longer produces a preview?)

What do you think how can I prove $T$ to be skew-symmetric?

 

[PeroK]

Don't you need the complex conjugate in there somewhere?

 

[fresh_42]

Don't you need the complex conjugate in there somewhere?

Where did you see complex numbers?

 

[pasmith]

Homework Statement:: Let $C(0.1)$ be the real linear space of all real functions continuous on $[0,1]$ with inner product
$$
\langle f , \rangle = \int_{0}^{1} f(t) ~g(t) dt $$
Let $V$ be the subspace of all $f$ such that
$$
\int_{0}^{1} f(t) dt = 0$$

Let $T: V \to C(0,1)$ be the integration operator defined by
$$
T (f(x)) = \int_{0}^{x} f(t) dt$$

This should be $$T(f)(x) = \int_0^x f(t)\,dt.$$

Prove that $T$ is skew-symmetric.
Relevant Equations:: $T$ will be skew-symmetric if
$$
\langle T(f), g \rangle = - \langle f, T(g)\rangle$$

$\langle T(f), g \rangle = \int_{0}^{1} \int_{0}^{x} f(t) dt ~ g(t) dt$
As $\int_{0}^{x} f(t) dt$ will be a function in $x$, therefore a constant w.r.t. $dt$, we have
$\langle T(f), g \rangle = \int_{0}^{x} f(t) dt ~ \int_{0}^{1} g(t) dt$

If $t$ is the dummy variable for the inner product integration, then you need a different symbol for the dummy variable in the definition of $T(f)(t)$. So $$T(f)(t) = \int_0^t f(s)\,ds$$ and the inner product with $g$ is $$\begin{split} \langle T(f),g \rangle &= \int_0^1 T(f)(t)g(t)\,dt \\ &= \int_0^1 \int_0^t f(s)\,ds\,g(t)\,dt. \end{split}$$ But it is simpler to define $F = T(f)$ and note that $F' = f$.

 

[PeroK]

Where did you see complex numbers?

I forgot an inner product could be real!

 

[Hall]

This should be $$T(f)(x) = \int_0^x f(t)\,dt.$$
If $t$ is the dummy variable for the inner product integration, then you need a different symbol for the dummy variable in the definition of $T(f)(t)$. So $$T(f)(t) = \int_0^t f(s)\,ds$$ and the inner product with $g$ is $$\begin{split} \langle T(f),g \rangle &= \int_0^1 T(f)(t)g(t)\,dt \\ &= \int_0^1 \int_0^t f(s)\,ds\,g(t)\,dt. \end{split}$$ But it is simpler to define $F = T(f)$ and note that $F' = f$.

I could get only this far:
$$
\langle T(f), g \rangle = \int_{0}^{1} \int_{0}^{t} f(s) ds g(t) dt$$
$$
\langle T(f), g \rangle = \int_{0}^{1} F(t) g(t) $$
Using integration by parts,
$$
\langle T(f), g \rangle = F(t) \int g(t) \big|_{0}^{1} - \int_{0}^{1} f(t) \int g(t) dt ~dt$$
$$
\langle T(f), g \rangle = F(t) G(t) \big|_{0}^{1} - \int f(t) G(t) dt $$

$$
\langle f, T(g) \rangle = \int_{0}^{1} f(t) \int_{0}^{t} g(s) ds ~dt$$

$$
\langle f, T(g) \rangle= G(t) \int f(t) dt \big|_{0}^{1} - \int_{0}^{1} g(t) F(t) dt $$
$$
\langle f, T(g) \rangle= F(t)~G(t) \big|_{0}^{1} - \int_{0}^{1} g(t) F(t) dt$$

 

[pasmith]

What do you know about $F(0) = \int_0^0 f(t)\,dt$ and $F(1) = \int_0^1 f(t)\,dt$, given that $f \in V$?

 

[Mark44]

I don't see that you used the given information that the domain of T is the set of functions whose integrals are zero.

Let $V$ be the subspace of all $f$ such that
$$\int_{0}^{1} f(t) dt = 0$$

Let $T: V \to C(0,1)$ be the integration operator defined by
$$
T (f(x)) = \int_{0}^{x} f(t) dt$$

(Can someone tell me why preview button no longer produces a preview?)

Preview works for me.

 

[Hall]

What do you know about $F(0) = \int_0^0 f(t)\,dt$ and $F(1) = \int_0^1 f(t)\,dt$, given that $f \in V$?

They are zero.

So, our expressions for inner product reduces:
$$
\langle T(f), g \rangle = - \int_{0}^{1} f(t) G(t) dt$$
And
$$
\langle f, T(g) \rangle = -\int_{0}^{1}g(t) F(t) dt$$

 

[Hall]

They are zero.

So, our expressions for inner product reduces:
$$
\langle T(f), g \rangle = - \int_{0}^{1} f(t) G(t) dt$$
And
$$
\langle f, T(g) \rangle = -\int_{0}^{1}g(t) F(t) dt$$

I think I got it, applying integration by parts once to the first inner product:
$$
\langle T(f), g \rangle = - \int_{0}^{1} f(t) G(t) dt$$
$$
\langle T(f) , g \rangle=
- \left( G(t) F(t) \big|_{0}^{1} - \int_{0}^{1} F(t) g(t) dt \right)$$
Again the first term will be zero, and we are remaining with
$$
\langle T(f), g \rangle = \int_{0}^{1} F(t) g(t) dt$$

Thus, we have
$$
\langle T(f), g\rangle = - \langle f, T(g)\rangle$$
hence, proving T is skew-symmetric.