Whether there exists a function $f(x,y)$ defined on $[0,1]\times(0,1]$ satisfies the following conditions: for any $x\in(0,1]$, $f(x,y)$ is decreasing with respect to $y$ and $\lim_{y\rightarrow0}f(x,y)=\log x$.
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Yes, for example $f(x,y) = \log x - y$
peterh
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@AndrewSalmon Also $\log x$ isn't. – peterh May 07 '14 at 07:55
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@user56927 What is its effect to the condition $\lim_{y\rightarrow0}f(x,y)=\log x$, if $\log x$ can't be defined? – peterh May 07 '14 at 08:02
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@PeterHorvath $\log x$ is not defined at $x = 0$, so your function is a function on $( 0,1 ] \times (0,1 ]$ but not on $[0,1] \times (0,1]$. – A.S May 07 '14 at 08:05
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@PeterHorvath This condition should only be established for any $x\in(0,1]$ – user56927 May 07 '14 at 08:05
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Pick any function $g(y)$ that is decreasing with $y$ with $g(0+) = 1$ and put $f(x,y) = \log(x+y) g(y)$.
For example $g(y) = \frac{\sin(y)}{y}$
Winther
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This is not decreasing; it's increasing, since $\log x$ is negative for $x < 1$ – A.S May 07 '14 at 07:56
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The point is that fixing $x < 1$, $\log x$ will be negative, so $\log (x) g(y)$ will be increasing, not decreasing. It's not clear whether $\log(x+y) g(x)$ is increasing or decreasing. – A.S May 07 '14 at 07:59
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You are right. I see! I answered a bit to fast. Need to modify it a bit then... – Winther May 07 '14 at 08:03
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What about something like this:
Let $h\, : \, (0,1] \to \mathbb{R}$ be any strictly increasing once continuously differentiable function with $\lim_{y \to 0^+} h(y) = 0$. Define $f\, : \, [0,1] \times (0,1] \to \mathbb{R}$ by $$f(x,y) = \begin{cases} \log(x) - h(y) & \text{ if } (x,y) \in (0,1] \times (0,1]\\ -h(y) & \text{ if } (x,y) \in \{0\} \times (0,1]\end{cases}$$
Then it is simple to verify that $\lim_{y\to 0^+} f(x,y) = \log(x)$ for all $x\in (0,1]$, and $f$ is decreasing in $y$ for fixed $x$ (i.e., $f_y < 0$ on $(0,1]$)
The wrinkle in this is that one should not expect any such example to be continuous.
Nicholas Stull
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