An "obvious" statement about a nonincreasing supremum Consider a nonnegative function $f(t,x): [0,\infty) \times [0,1] \rightarrow [0, \infty)$.  Suppose we have the following property:
$$ \mbox{ If } ~~~~~~~~~~f(t,y) > \frac{1}{2} \sup_{x \in [0,1]} f(t,x) ~~~~~~~\mbox{ then}~~~~~~~ \frac{d}{dt} f(t,y) < 0 $$ In words, if any $y$ has $f(t,y)$ at least half as big as the largest $f(t,\cdot)$, then $f(t,y)$ is decreasing.
It seems very natural to guess that $$\sup_{x \in [0,1]} f(t,x) \mbox{ is nonincreasing in } t$$ This seems obvious - the ``top half'' of $f(t, \cdot)$ is always decreasing, so how can the supremum increase? But I don't see how to show this mathematically. Thus my question is whether this is true. 
Some comments:


*

*The above implicitly assumes that for each $x$, $f(t,x)$ is differentiable with respect to $t$. To make sure nothing weird happens at $t=0$, let us also assume that the function $f(t,x)$ is a continuous function of $t$ for each $x$.  

*If it helps, we can further assume that its derivative $f_t(t,x)$ is a continuous function of $t$. 

*Note that there is no assumption of differentiability or even continuity with respect to $x$. Furthermore, there is no assumption that the suprema in question must remain finite. 
 A: Counterexample provided by Daniel Fischer in a comment:
$$f(t,x) = \begin{cases} e^{-t} ,\quad  &x = 0\\ t/x ,\quad  &x > 0\end{cases}$$
Observe that  $\sup_{x\in [0,1]}f(t,x) = 1$ when $t=0$ and $\sup_{x\in [0,1]}f(t,x) = \infty$ when $t>0$.
Indeed, for each fixed $x$, $f(t,x)$ is a very nice function of $t$. The condition of having negative $t$-derivative holds at $t=0$, where only $e^{-t}$ is involved in it. It also holds for positive $t$, simply because the supremum is infinite while the functions are finite; so none of the functions have $f(t,y)>\frac12 \sup_{x\in [0,1]}f(t,x)$. 
A: I'll assume "nonincreasing" means "not-necessarily-stricly decreasing". In this case, it is possible to construct a counterexample.
For each $n$, let $\phi_n:[0,\infty)\to[0,\infty)$ be a function satisfying:


*

*$\phi_n(t)=t^{-1}$ for $t\in[3^{-n},\infty)$,

*$\phi_n(t)\leq\min\{t^{-1},3^n+1\}$ for all $t$,

*$\phi_n(0)=0$,

*$\phi_n$ is strictly decreasing on $[3^{-n-1},3^{-n}]$,

*$\phi_n$ is smooth.


Such functions exist by standard bump function arguments.
Now, define $$f(t,x)=\begin{cases}e^{-t};&x=0,\\\phi_n(t);&x=\frac1n,\text{ where $n\in\mathbb N$,}\\0;&\text{otherwise.}\end{cases}$$
To simplify notation, let $s_t$ be the supremum under consideration at $t$. Our definition implies that $s_t=t^{-1}$ for $t>0$ and $s_0=1$. (In particular, $s_t$ is always finite and not nonincreasing.)
The function $f$ satisfies the condition on the derivative at $t=0$, since $e^{-t}$, i.e. the only function with value greater than $\frac12s_0$ at $0$, is strictly decreasing everywhere. For $t>\frac13$, each function $f(\cdot,x)$ is either strictly decreasing or zero, so the condition holds. For $t\in(0,\frac13]$, there is a $n\in\mathbb N$ such that $t\in(3^{-n-1},3^{-n}]$. Here, it is only necessary to consider the functions $\phi_1,\ldots,\phi_{n-1}$, since the others are either less than $\frac12 t^{-1}$ or strictly decreasing by assumption. But none of these takes values greater than $3^{n-1}+1<\frac12s_t$.
Therefore, $f$ is a counterexample.
Remark. If we assume that the supremum is finite over $[0,\infty)\times[0,1]$, the conclusion might change.
