$\int_0^x g(x-a)f(a)da=\int_0^x f(a)da$ - Please help to explain the error

Question

Let $I$ be an integral function s.t.

$$I(f(x))= \int_0^x f(a)\,da.$$

Let $g$ be a function s.t. $g(0)=1$ and $g_x(a):=g(x-a)$.

I have

\begin{align} \int_0^x g_x(a)f(a)\,da& =I(g_x(x)f(x)) \\[8pt] & =I(f(x)) \\[8pt] & =\int_0^x f(a)\,da \end{align}

Ie:

If $g(0)=1$, then $\int_0^x g(x-a)f(a)\,da=\int_0^x f(a)\,da$.

I notice this is wrong, but 'd like so much any explanation. I thought I cannot get the first equality because $g_x$ also depends on $x$. If this is the explanation, I'd like to know if there some way of write $g(x)\ast f(x)=\int_0^x g_x(a)f(a)\,da$ as an integral function.

Many thanks!

Answer 1

I presume this is an instance of confusion coming from a bad notation. The notation should be $I(f)(x)=\int_0^x f(a)\,da$, and therefore $I(g_yf)(x)=\int_0^x g(y-a)f(a)\,da$. I presume your $I(g_x(x)f(x))$ is supposed to stand for the function $x\mapsto I(g_xf)(x)$.