Prove that a convex function minus a constant is convex

Question

Let $f : \mathbb{R}^d \rightarrow \mathbb{R}$, be a convex function, so $\forall x, y \in \mathbb{R}^d$ $$f(tx+(1-t)y) \leq tf(x)+(1-t)f(y)$$ $t \in[0,1]$. How do you show that $s(x) = f(x)-k$ also is convex ($k\in\mathbb{R}$)?

I only get: $$s(tx+(1-t)y) \leq tf(x)+(1-t)f(y) -k =ts(x)+(1-t)s(y)+k$$

Answer 1

I think you just have an arithmetic error; namely,

$$s(tx + (1-t)y) \leq tf(x) + (1-t)f(y) - k = t(f(x)-k) + (1-t)(f(y)-k).$$