Two Integrals Evalued at the Same Values

Question

If two integrals evaluated at the same points are equal, what can be said about them?

I am trying to prove or disprove that the $\phi(f)=\int_{0}^{4}f(x) dx$ where $\phi:$ F $\rightarrow \mathbb{R}$, where F is additive group of all continuous functions mapping $\mathbb{R}$ to $\mathbb{R}$ is an isomorphism.

For one-to-one: Let $F(x)=\int f(x)dx$ and $G(x)=\int g(x)dx$ and $\phi(F(x))=\phi(G(x))$. Then $\int_{0}^{4} f(x)dx = \int_{0}^{4} g(x)dx$. Then $F(x)\Big|_0^4 = G(x)\Big|_0^4$.

This is where I get stuck.

Answer 1

It isn't injective since the continuous functions $f(x)=-x+4$ and $g(x)=x$ are different functions that have the same integral on $[0,4]$

Answer 2

Injectivity fails: for example, consider the linear function with graph that passes through the points $(0,-2)$ and $(4,2)$ (i.e. the function $f(x) = x-2$) and the zero function (i.e. $g(x) = 0$). Both functions are continuous on $\mathbb{R}$, but $$ \varphi(f) = \int_{0}^{4} f(x)\,\mathrm{d}x = \int_{0}^{4} x-2\,\mathrm{d}x = \left[ \frac{x^2}{2} - 2x \right]_{x=0}^{4} = 0,$$ and $$ \varphi(g) = \int_{0}^{4} g(x)\,\mathrm{d}x = \int_{0}^{4} 0\, \mathrm{d}x = 0. $$ Indeed, simply noting that $\varphi(f) = 0$ is sufficient to show that it is not an isomorphism as the kernal of $\varphi$ is non-trivial.

Answer 3

This isn't an isomorphism. There are real valued continous functions $f(x)$ which are not identically $0$; however,

$$\int_{0}^4 f(x) dx = 0$$

Take $f(x) = sin(2 \pi x)$ for instance.