# Proving the partial derivative of a strictly decreasing function is finite everywhere

Suppose I have a function $F:[0,1] \times [0,\infty) \rightarrow (0,1]$ of two variables $(s,t)$ that satisfies:

(1) $F(s,0)=1$ for each $s \in [0,1]$.

(2) $\lim_{t \rightarrow \infty} F(s,t)=0$ for all $s \in [0,1]$.

(3) For each fixed $s \in [0,1]$, the function $t\mapsto F(s,t)$ is strictly decreasing.

Consider the function $G:[0,1] \times [0,\infty) \rightarrow \mathbb{R}$ defined by $G(s,t)=\frac{\partial F}{\partial s}(s,t)$.

Is it necessarily the case that $\sup_{(s,t) \in [0,1] \times [0,\infty)} \left|G(s,t)\right| \lt \infty$?

If not, how would I go about constructing a counter-example?

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The assumptions (1)–(3) don't even imply that $\partial F/\partial s$ exists (the functions $t \mapsto F(s,t)$ for different values of $s$ can be chosen independently of each other). –  Hans Lundmark Jan 7 '11 at 16:01

Hint: Exponentials are handy for things like this because they are easy to take derivatives of and go nicely to zero if the exponent gets large and negative. Let $F(s,t)=\exp(g(s,t))$ So you want to construct a function $g(s,t)$ such that $g(s,0)=0, \lim_{t \rightarrow \infty} g(s,t)=-\infty, \lim_{s \rightarrow 0}\frac{\partial g(s,t)}{\partial s}=-\infty$. Something like $g(s,t)=-\frac{t}{h(s)}$ with $h$ chosen properly will do the trick.