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

Hall
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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?
 
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Don't you need the complex conjugate in there somewhere?
 
PeroK said:
Don't you need the complex conjugate in there somewhere?
Where did you see complex numbers?
 
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Hall said:
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 <br /> 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 <br /> T(f)(t) = \int_0^t f(s)\,ds and the inner product with g is <br /> \begin{split}<br /> \langle T(f),g \rangle &amp;= \int_0^1 T(f)(t)g(t)\,dt \\<br /> &amp;= \int_0^1 \int_0^t f(s)\,ds\,g(t)\,dt. \end{split}<br /> But it is simpler to define F = T(f) and note that F&#039; = f.
 
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fresh_42 said:
Where did you see complex numbers?
I forgot an inner product could be real!
 
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pasmith said:
This should be <br /> 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 <br /> T(f)(t) = \int_0^t f(s)\,ds and the inner product with g is <br /> \begin{split}<br /> \langle T(f),g \rangle &amp;= \int_0^1 T(f)(t)g(t)\,dt \\<br /> &amp;= \int_0^1 \int_0^t f(s)\,ds\,g(t)\,dt. \end{split}<br /> But it is simpler to define F = T(f) and note that F&#039; = 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,
$$
\langle T(f), g \rangle = F(t) \int g(t) \big|_{0}^{1} - \int_{0}^{1} f(t) \int g(t) dt ~dt$$
$$
\langle T(f), g \rangle = F(t) G(t) \big|_{0}^{1} - \int f(t) G(t) dt $$

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

$$
\langle f, T(g) \rangle= G(t) \int f(t) dt \big|_{0}^{1} - \int_{0}^{1} g(t) F(t) dt $$
$$
\langle f, T(g) \rangle= F(t)~G(t) \big|_{0}^{1} - \int_{0}^{1} g(t) F(t) dt$$
 
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?
 
I don't see that you used the given information that the domain of T is the set of functions whose integrals are zero.
Hall said:
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
$$
T (f(x)) = \int_{0}^{x} f(t) dt$$

Hall said:
(Can someone tell me why preview button no longer produces a preview?)
Preview works for me.
 
pasmith said:
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
$$
\langle f, T(g) \rangle = -\int_{0}^{1}g(t) F(t) dt$$
 
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Hall said:
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.
 
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