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For functions $f,g:X\to \mathbb{R}$ or $\mathbb{C}$, their pointwise multiplication is the function $fg$ with $(fg)(x)=f(x)g(x)$. Both values are evaluated at the same input before multiplication. If $f(x)=x+1$ and $g(x)=x^2$, then $(fg)(2)=3\cdot 4=12$ and $(fg)(x)=x^3+x^2$.

Together with pointwise addition, this makes the continuous functions on a topological space into an algebra over a field. Closure properties still need checking for other function spaces: two L2 space functions need not have an L2 space product. For example, $f(x)=g(x)=x^{-1/3}$ on $(0,1)$ have integrable squares, but $(fg{)}^2=x^{-4/3}$ is not integrable. The ambient function space matters.

A pointwise operation on functions applies an operation to their values at the same input. If $f,g:X\to Y$ and $\star :Y\times Y\to Y$ is a binary operation, the induced operation on the function space is defined byThe common domain of a function makes the two inputs available at each $x$; the closure of the operation in the codomain makes the resulting value an element of $Y$ again.

For real-valued functions, ordinary addition and multiplication induce pointwise addition and pointwise multiplication. If $f(x)=x+1$ and $g(x)=x^2$, then $(f+g)(x)=x^2+x+1$ and $(fg)(x)=x^2(x+1)$. Function composition instead feeds one output into another rule: $(f\circ g)(x)=x^2+1$ in this example.

Algebraic identities pass to the pointwise construction by evaluation. For instance, if the operation on $Y$ is associative, then $((f\star g)\star h)(x)=(f(x)\star g(x))\star h(x)=f(x)\star (g(x)\star h(x))$ for every $x$, proving equality of the two functions. This explains how algebraic structure on values induces structure on a function space.