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Homework Help: Analysis Problem (proving differentiability at a point)

  1. Oct 26, 2004 #1

    AKG

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    Need some hints on how to go about doing this:

    [tex]f(x, y)=\left\{\begin{array}{cc}\frac{x^4 + y^4}{x^2 + y^2},&\mbox{ if }
    (x, y)\neq (0,0)\\0, & \mbox{ if } (x, y) = 0\end{array}\right.[/tex]

    Show that f is differentiabile at [itex](0, 0)[/itex].

    I've tried a number of things, too ugly and not worth writing down here (all got me nowhere). As far as I can tell, I want to show that some linear transformation [itex]\lambda \ :\ \mathbb{R}^2 \rightarrow \mathbb{R}[/itex] satisfies the equation:

    [tex]\lim _{h \rightarrow 0} \frac{|f(0 + h) - f(0) - \lambda (h)|}{|h|} = 0[/tex]

    [tex]\lim _{h \rightarrow 0} \frac{|f(h) - \lambda (h)|}{|h|} = 0[/tex]

    [tex]\lim _{h \rightarrow 0} \left ( \frac{h_1^4 + h_2^4}{|h|^3} - \frac{\lambda (h)}{|h|} \right ) = 0[/tex]

    Note that [itex]h = (h_1, h_2) \in \mathbb{R}^2[/itex]

    Now, if we let [itex]\mu = -\lambda[/itex], then we have, and [itex]h_2 = 0[/itex], then we have:

    [tex]\lim _{h \rightarrow 0} \left ||h_1| + \frac{\mu (h)}{|h_1|} \right |[/tex]

    [tex]\leq \lim _{h \rightarrow 0} |h_1| + \left | \frac{\mu (h)}{|h_1|} \right |[/tex]

    We know that there exists some real M > 0 such that [itex]|\mu (v)| \leq M|v|\ \forall \ v \in \mathbb{R}^2[/itex], so:

    [tex]\leq \lim _{h \rightarrow 0} |h_1| + M = M[/tex]

    So, if the function is differentiable at 0, then M = 0, so the linear transformation [itex]\mu[/itex] is the zero transformation, so the derivative at 0 is the zero transformation. Tell me if I've made a mistake so far, because, if not, I think I can prove that it's also not the zero transformation. If I try to picture the graph, I think it should be zero. But I want to show that there exists some transformation such that :

    [tex]\lim _{h \rightarrow 0} \frac{|f(0 + h) - f(0) - \lambda (h)|}{|h|} = 0[/tex]

    holds, and when I try [itex]\lambda = 0[/itex], I just can't seem to evaluate the limit right (or rather, prove that it will evaluate to zero). However, I do seem to be able to show that it will be greater than zero, meaning that, if there is a derivative, the zero transformation is not it (contrary to what I just showed above with M = 0). So, clearly, I'm stuck. Any help would be appreciated.

    EDIT: Actually, I think I can do the proof, here's what I have.

    Assuming what I've done is right so far, and [itex]\lambda = \mu = 0[/itex] (the zero transformation), then I need to prove:

    [tex]L = \lim _{h \rightarrow 0} \frac{|f(h)|}{|h|} = 0[/tex]

    [tex]L = \lim _{h \rightarrow 0} \frac{|h_1^4 + h_2^4|}{(h_1^2 + h_2^2)^{3/2}}[/tex]

    [tex]= \lim _{h \rightarrow 0} \frac{|(h_1^2 + h_2^2)^2 - 2h_1^2h_2^2|}{(h_1^2 + h_2^2)^{3/2}}[/tex]

    [tex]= \lim _{h \rightarrow 0} \left ||h| - \frac{2h_1^2h_2^2}{(h_1^2 + h_2^2)^{3/2}} \right |[/tex]

    [tex]= 2\lim _{h \rightarrow 0} \frac{h_1^2h_2^2}{(h_1^2 + h_2^2)^{3/2}}[/tex]

    [tex]= 2\lim _{h \rightarrow 0} \left (|h|\frac{h_1^2h_2^2}{(h_1^2 + h_2^2)^2} \right )[/tex]

    [tex]= 2\lim _{h \rightarrow 0} |h| \left (\frac{h_1h_2}{h_1^2 + h_2^2} \right )^2[/tex]

    Now, consider the function [itex]g(z) = z + \frac{1}{z}[/itex] for positive [itex]z \in \mathbb{R}[/itex]. Simple analysis shows that g reaches a minimum at 2, so:

    [tex]z + \frac{1}{z} \geq 2[/tex]

    Now, let [itex]\frac{|h_1|}{|h_2|} = z[/itex]. Now, if either component of h is zero, we could have proven that the L = 0 long ago, so for the case where neither is zero, we can assign z as we have above. Now, we have:

    [tex]\frac{|h_1|}{|h_2|} + \frac{|h_2|}{|h_1|} \geq 2[/tex]

    [tex]h_1^2 + h_2^2 \geq 2|h_1||h_2|[/tex]

    [tex]\frac{1}{2} \geq \frac{|h_1||h_2|}{h_1^2 + h_2^2}[/tex]

    [tex]\frac{1}{4} \geq \left ( \frac{h_1h_2}{h_1^2 + h_2^2} \right )^2[/tex]

    [tex]2|h|\frac{1}{4} \geq 2|h|\left ( \frac{h_1h_2}{h_1^2 + h_2^2} \right )^2[/tex]

    So:

    [tex]L \leq \frac{1}{2}\lim _{h \rightarrow 0} |h| = 0[/tex]

    And so the proof is done. Did I do it right?
     
    Last edited: Oct 26, 2004
  2. jcsd
  3. Oct 26, 2004 #2

    AKG

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    Originally, I had trouble with this question (which is why I posted it here). However, I think I got it (see edit in original post) so if anyone could check and make sure I did it right, that would be great. Thanks.
     
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