$$ \alpha\leq f(t)\leq\beta,\quad \forall t\in[0,1]\\ f(0)=f_0,\quad f(1)=f_1 $$ where $f_0$ and $f_1\in[\alpha,\beta]$.
Is the feasible set of $f$ convex in $f$?
Thanks in advance!
$$ \alpha\leq f(t)\leq\beta,\quad \forall t\in[0,1]\\ f(0)=f_0,\quad f(1)=f_1 $$ where $f_0$ and $f_1\in[\alpha,\beta]$.
Is the feasible set of $f$ convex in $f$?
Thanks in advance!
Of course it is convex, for if $f_1,f_2,\dots f_n$ is any finite collection of functions in your "feasible set", and $\sum_{i=1}^n\lambda_i=1$ for some $\lambda_i\geq 0$, then the function $g(t)=\sum_{i=1}^n\lambda_if_i(t)$ clearly satisfies all the conditions: $$\alpha\leq g(t)\leq \beta\enspace\forall t\in[0,1]\enspace\hbox{and} \enspace g(0)=f_0,\enspace g(1)=f_1$$ and so $g(t)$ also belongs to the "feasible set", so the latter is a convex set.