Is there any property of Hypergeometric function due to which we can prove following equality? $$-\frac{u^2}{2}+\frac{u^2}{2}\left[\,_2F_1\left(m;-\frac{2}{k};1-\frac{2}{k};\frac{a}{u^k}\right)\right]=f(\infty)-f(u)$$ where $f(x)$ is defined as follows $$f(x)=\frac{x^2}{2}-\frac{x^2}{2}\,_2F_1\left(-\frac{2}{k},m;1-\frac{2}{k};-\frac{a}{x^k}\right).$$ where $a,u,m$ are positive values and $k>2$. In my thinking if the limit $\infty$ is zero and if we swap the positions of $-\frac{2}{k}$ and $m$ (in the first two arguments of Hypergeometric function then the answer is ok, but I am not sure how to prove these things). Thanks in advance.
On Limit $x\to \infty$:
We can write $f(x)$ as $$f(x)=\frac{x^2}{2}-\frac{x^2}{2}\sum_{n=0}^{\infty}\frac{\left(-\frac{2}{k}\right)_n(m)_n}{(1-\frac{2}{k})_nn!}\left(-\frac{a}{x^k}\right)^n$$ now if we expand it fof $n$ then the first term in the summation will cancel out with $\frac{x^2}{2}$ and the limit can be applied on the remaining terms and this will make sure that $f(\infty)=0$ because $n>0,k>0$.
Since $_2F_1(a,b;c;z) =\sum\limits_{n=0}^{\infty}\dfrac{(a)_n(b)_n}{(c)_n}\dfrac{z^n}{n!} $, $_2F_1(a,b;c;z) = _2F_1(b,a;c;z) $ so swapping $a$ and $b$ is OK.
(added after OP's comment)
Pfaff's transformation is $_2F_1(a,b;c;z) = (1-z)^{-a}\,_2F_1(a, c-b;c;\frac{z}{z-1}) $ or $(1-z)^a\,_2F_1(a,b;c;z) =\, _2F_1(a, c-b;c;\frac{z}{z-1}) $. As $z \to 1$, $\frac{z}{z-1} \to \infty$, so that, using your notation, $\, _2F_1(a, c-b;c;\infty) =\lim_{z \to 1}(1-z)^a\,_2F_1(a,b;c;1) $.
Therefore, if $a > 0$, $\, _2F_1(a, c-b;c;\infty) =0$.