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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?

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?