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

• Hall
In summary, we have shown that the integration operator T on a subspace V of real functions is skew-symmetric, as its inner product satisfies the property ##\langle T(f), g \rangle = - \langle f, T(g)\rangle## for all ##f,g \in V##. This is due to the fact that the integrals of functions in V are zero, resulting in a simplified expression for the inner product.
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?

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?

PeroK
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 $$T(f)(x) = \int_0^x f(t)\,dt.$$

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 $F' = f$.

Delta2, PeroK and Hall
fresh_42 said:
Where did you see complex numbers?
I forgot an inner product could be real!

Delta2
pasmith said:
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,
$$\langle T(f), g \rangle = F(t) \int g(t) \big|_{0}^{1} - \int_{0}^{1} f(t) \int g(t) dt ~dt$$
$$\langle T(f), g \rangle = F(t) G(t) \big|_{0}^{1} - \int f(t) G(t) dt$$

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

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

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
$$\langle f, T(g) \rangle = -\int_{0}^{1}g(t) F(t) dt$$

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.

Last edited:
Delta2

1. How do you prove that a matrix is skew-symmetric?

To prove that a matrix ##T## is skew-symmetric, you must show that ##T^T = -T##, where ##T^T## is the transpose of ##T##. This means that the entries of ##T## must satisfy the condition ##t_{ij} = -t_{ji}## for all ##i## and ##j##.

2. What is an inner product?

An inner product is a mathematical operation that takes two vectors and returns a scalar value. It is often denoted by ##\langle \cdot, \cdot \rangle## and satisfies properties such as symmetry, linearity, and positive definiteness.

3. How is an inner product related to integration?

In some cases, an inner product can be thought of as a generalization of the concept of integration. This is because both operations take two objects and return a scalar value. However, an inner product can be defined on more general spaces than just functions, such as matrices or vectors.

4. Can you give an example of an inner product?

One example of an inner product is the dot product, which takes two vectors and returns the sum of the products of their corresponding entries. For example, the dot product of ##[1,2,3]## and ##[4,5,6]## is ##1\cdot4 + 2\cdot5 + 3\cdot6 = 32##.

5. Why is it important to prove that an inner product is an integration?

Proving that an inner product is an integration is important because it allows us to use the tools and techniques of integration to study and analyze the properties of inner products. This can be especially useful in applications where inner products are used, such as in quantum mechanics or signal processing.

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