I am currently reading Calculus of Variations by I.M Gelfand and S. V. Fomin.
One of the lemmas in the book states
If $\alpha(x)$ is continuous in $[a,b]$, and if
$$\int_a^b \alpha(x)h'(x)=0$$
for every function $h(x)\in \mathcal{D}_1(a,b)$ such that $h(a)=h(b)=0$, then $\alpha(x)=c$ for all $x$ in $[a,b]$, where $c$ is a constant
Now $\mathcal{D}_1(a,b)$ they define as a normed linear space and $\alpha(x)$ as a fixed function
In the proof they define $h(x) = \int_a^x [\alpha(\xi)-x]d\xi$ $(1)$ so that $h(x)$ belongs to $\mathcal{D}_1(a,b)$ and use this later to derive the result $$\int_a^b [\alpha(x)-c]h'(x)dx = \int_a^b [\alpha(x)-c]^2dx$$
My Question:
Does this mean since $\alpha(x)$ is a "fixed function" that $(1)$ can be expressed as $$h(x) = \int_a^x [\alpha(\xi)-x]d\xi = (\alpha(x)-c)x-(\alpha(a)-c)a$$ Which would make its derivative $$h'(x) = \alpha(x)-c$$ Which is what we need to derive the above result on the proof
Additional Question:
- Does this mean that fixed functions can be treated as a constant would be in normal calculus
- If so do we only substitute the limits of the integral into them when working with integrals
- Finally where can I find a better definition of them or am I just miss interpreting the nomenclature