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I am trying to figure out differentiability of $\mathrm{max}(x, y)$. Intuitively, it should not be differentiable at $x=y$, since it changes direction "non-smoothly" at those points.

I can not, however, find a way to prove this algebraically.

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Hint: Write

$$\operatorname{max}(x,y) = \frac{x + y + |x-y|}{2}$$

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Take any point $(a,a)$ and prove that the partial derivative does not exist. The partial derivative is $\lim_{x\rightarrow a} \frac{f(x,a)-f(a,a)}{x-1}$. The left limit for this is $$\lim_{x\uparrow a}\frac{\max\{x,a\}-a}{x-a}=\lim_{x\uparrow a}\frac{a-a}{x-a}=0$$ while the right limit is $$\lim_{x\downarrow a}\frac{\max\{x,a\}-a}{x-a} = \lim_{x\downarrow a}\frac{x-a}{x-a}=1$$

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Let $f(x,y)=\max \{x,y\}$. Now look at $f$ on the section $y=-x$: $$ f(x,-x)=\max \{x,-x\}=|x|. $$ This function is not differentiable as a function of the variable $x$, so $f$ can't be differentiable.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \partiald{\max\pars{x,y}}{x} = \half\bracks{1 + \sgn\pars{x - y}} $$

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    $\begingroup$ Formula valid for... $x\ne y$. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 10 '14 at 14:06
  • $\begingroup$ @Martín-BlasPérezPinilla Yes. I agree. Thanks. $\endgroup$ – Felix Marin Feb 10 '14 at 14:18
  • $\begingroup$ @Martín-BlasPérezPinilla Consider the $\large{\rm sgn}$ function as a distribution and it will make sense. $\endgroup$ – Felix Marin Feb 10 '14 at 23:36
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    $\begingroup$ Distributions? These seem wildly out of the scope of the question, don't you think? $\endgroup$ – Did Feb 11 '14 at 16:10

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