Prove that set of all functions from R to R has multiplication distributive:Question?

Suppose $\displaystyle \mathfrak{F}(\mathbb{R})$ represents the set of all the functions from $\displaystyle \mathbb{R}$ to $\displaystyle \mathbb{R}$;that is, the set of all real-valued functions of a set of a real variable. Prove that multiplication is distributive over addition for $\displaystyle \mathfrak{F}(\mathbb{R})$.

What this means is:

Let $\displaystyle f$, $\displaystyle g$ and $\displaystyle h$ are three functions.

We have to prove that(part 1) $\displaystyle f(g + h) = fg + fh$ and (part 2) $\displaystyle (g + h)f = gf + hf$

**Proof:**

Now from calculus we know that for two functions $\displaystyle a$ and $\displaystyle b$:

$\displaystyle [a + b](x) = a(x) + b(x) \text{ for every real number x }$...........(1)

$\displaystyle \text{and }[ab](x) = a(x)b(x) \text{ for every real number x }$......(2)

Proof of part 1:

$\displaystyle \begin{align*}(f[g+h])(x) =& [fg+fh](x) \\ =& fg(x) + fh(x)\text{.........by(1)} \end{align*}$

Proof of part 2:

$\displaystyle \begin{align*}([g+h]f)(x) =& [gf+hf](x) \\ =& gf(x) + hf(x)\text{.........by(1)} \end{align*}$

So $\displaystyle \mathfrak{F}(\mathbb{R})$ has multiplication distributive over addition.[Q.E.D]

The steps I've taken to prove this are they correct? Am I right to prove it like this? Can anyone kindly confirm?

Re: Prove that set of all functions from R to R has multiplication distributive:Quest

I don't think its quite that simple.

This set you describe forms a vector space, where the vectors are functions f,g,etc...

it doesn't make sense to multiply vectors. meaning [f[g+h]](x) means add the vectors g + h, multiply by vector f and evaulate it at a point x from the field $\displaystyle \mathbb{R}$.

instead for part 1. h(x)[(f+g)(x)] = h(x)[f(x)+g(x)] = h(x)f(x) + h(x)g(x)

Since by the vector space axioms (f+g)(x) = f(x) + g(x). so the above holds.

The same problem with #2,

It might seem as though its a small problem but the brackets and paranthesis can make the statement mean something completely different.

Re: Prove that set of all functions from R to R has multiplication distributive:Quest

Quote:

Originally Posted by

**x3bnm** Now from calculus we know that for two functions $\displaystyle a$ and $\displaystyle b$:

$\displaystyle [a + b](x) = a(x) + b(x) \text{ for every real number x }$...........(1)

$\displaystyle \text{and }[ab](x) = a(x)b(x) \text{ for every real number x }$......(2)

You are saying this as if these are theorems, but, in fact, these are definitions of addition and multiplications on functions.

Quote:

Originally Posted by

**x3bnm** Proof of part 1:

$\displaystyle \begin{align*}(f[g+h])(x) =& [fg+fh](x) \\ =& fg(x) + fh(x)\text{.........by(1)} \end{align*}$

You have justified the second equality, but not the first one.

Quote:

Originally Posted by

**jakncoke** This set you describe forms a vector space, where the vectors are functions f,g,etc...

it doesn't make sense to multiply vectors.

Then it is not just a vector space (but something like an algebra over a field). However, I am not sure talking about vector spaces is needed here; the problem is easy as is.

Quote:

Originally Posted by

**jakncoke** meaning [f[g+h]](x) means add the vectors g + h, multiply by vector f and evaulate it at a point x from the field $\displaystyle \mathbb{R}$.

But you said that it doesn't make sense to multiply vectors (i.e., functions).

Quote:

Originally Posted by

**jakncoke** instead for part 1. h(x)[(f+g)(x)] = h(x)[f(x)+g(x)] = h(x)f(x) + h(x)g(x)

Since by the vector space axioms (f+g)(x) = f(x) + g(x). so the above holds.

By what vector space axiom? This is not multiplication by a scalar; this is application of a function to an argument!

Here is a proof. We need to show that h(f + g) = hf + hg. By definition of (pointwise) equality of functions, this means [h(f + g)](x) = [hf + hg](x) for every x. So fix an arbitrary x.

$\displaystyle \begin{align*}[h(f + g)](x)&=h(x)[(f+g)](x) && \text{by (2)}\\ &=h(x)(f(x)+g(x)) &&\text{by (1)}\\ &=h(x)f(x)+h(x)g(x) &&\text{by distributivity of multiplication on }\mathbb{R}\\ &=[hf](x)+[hg](x)&& \text{by (2)}\\ &=[hf+hg](x)&& \text{by (1)}\end{align*}$

Re: Prove that set of all functions from R to R has multiplication distributive:Quest

Quote:

Originally Posted by

**emakarov** You are saying this as if these are theorems, but, in fact, these are definitions of addition and multiplications on functions.

You have justified the second equality, but not the first one.

Then it is not just a vector space (but something like an algebra over a field). However, I am not sure talking about vector spaces is needed here; the problem is easy as is.

But you said that it doesn't make sense to multiply vectors (i.e., functions).

By what vector space axiom? This is not multiplication by a scalar; this is application of a function to an argument!

Here is a proof. We need to show that h(f + g) = hf + hg. By definition of (pointwise) equality of functions, this means [h(f + g)](x) = [hf + hg](x) for every x. So fix an arbitrary x.

$\displaystyle \begin{align*}[h(f + g)](x)&=h(x)[(f+g)](x) && \text{by (2)}\\ &=h(x)(f(x)+g(x)) &&\text{by (1)}\\ &=h(x)f(x)+h(x)g(x) &&\text{by distributivity of multiplication on }\mathbb{R}\\ &=[hf](x)+[hg](x)&& \text{by (2)}\\ &=[hf+hg](x)&& \text{by (1)}\end{align*}$

you are right, i was thinking of something else and incorrectly at that .

Re: Prove that set of all functions from R to R has multiplication distributive:Quest

Thanks emakarov and jakncoke. I understand now.