Find the values of the real parameter $a$ such that the function $f:[0,1] \to R, f(x)=x^2-|x-a|$ is a convex function. It is clear that if $a=0$ or $a=1$ the function will be the restriction of a quadratic function, thus convex.
And intutively f is not convex when $ a \in (0,1)$ because in that case the function would have "a concavity" around the point $a$. but i would be interested in a rigorous approach in this case.
You said:
That is correct, and holds even for $a \le 0$ and for $a \ge 1$.
Then you said:
For a rigorous approach we can compute $$ \frac 12 \bigl( f(a-h) + f(a+h)\bigr) - f(a) $$ which should be $\ge 0$ for a convex function. But if $0 < a < 1$ and $0 < h < \min(a, 1-a)$ (so that all terms are defined) this expression evaluates to $$ h^2-2h = (1-h)^2 - 1 < 0 \, , $$ so that $f$ is not convex.
Alternatively: If $f$ were convex then $$ f(x) \le (1-x)f(0) + x f(1) = (1-x)(-a) + xa = -a + 2ax $$ for all $x \in [0, 1]$, this is violated at $x=a$ with $f(a) = a^2$.
Yet another approach for the case $0 < a < 1$ would be to observe that $$ f'(x) = \begin{cases} 2x + 1 & \text{for } 0 \le x < a \\ 2x - 1 & \text{for } a < x \le 1 \\ \end{cases} $$ which also demonstrates that $f$ is not convex on $[0, 1]$, because a convex function has a right (and left) derivative at every point, and the right (and left) derivative is monotonically increasing.