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Consider the problem of minimizing the functional $$ J(y) = \int_0^1 y'(x)^2(1-y'(x)^2)dx $$ with the end-points condition $y(0) = y(1) = 0$. Is the curve $y \equiv 0$ a weak minimum or a strong minimum of $J$?

Clearly, $y$ is a weak minimum. How can I make a conclusion about $y$ being a strong minimum or not?

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    Can any other function in your function space have $J[y]=0$? Note that the identity of the function space is important here: if for instance it is enough for $y$ to be just absolutely continuous, then I can make up another function with $J[y]=0$ (take a function that has slope $1$ half the time and slope $-1$ the other half of the time). – 2017-02-23
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    It would appear the answer is no, so $y$ is a strong minimum? I think I am supposed to make a conclusion regarding the differing norms for weak and strong. Is there a case here for that? – 2017-02-23
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    You need to say what your space is (including both what is in it and what its norm is). The maximal domain of $J$ is the Sobolev space $W^{1,4}_0((0,1))$, but the answer to your question may not be the same as the answer on $W^{1,4}_0$. – 2017-02-23
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    It is unstated, but the function space is typically $C^1[0,1]$ and $\|f\|_{1} = \|f\|_{\infty} + \|f'\|_{\infty}$ and $\|f\|_{\infty} = \max_{x\in[0,1]}|f(x)|$. Weak min corresponds to the $1$-norm and strong min to the $\infty$-norm. – 2017-02-23
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    Try looking at a mollification of $f_\epsilon(x)=\epsilon(1-2|x-1/2|)$ when $\epsilon>0$. This will have small $C^1$ norm and a small value of $J$. – 2017-02-23
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    Hm, actually I think I screwed that up, because the point of the example was to make $y'$ be nearly $\pm 1$ at all times but I actually made it $\pm 2 \epsilon$ which isn't desirable here. Can you double check on what norms you're supposed to use, and what functions are permitted? This seems like a subtle problem actually... – 2017-02-23
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    @Ian $C^1[0,1]$ functions are permitted with norm $\|f\|_{\infty} = \max_{x\in[0,1]}|f(x)|$ and $\|f\|_1 := \|f\|_{\infty} + \|f'\|_{\infty}$ where weak minimums are with respect to $\|\cdot\|_1$ and strong minimums are with respect to $\|\cdot\|_{\infty}$. – 2017-02-23
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    OK. I think it's standard, but just to check, can you write out the definition of a minimum of a function with respect to a norm? – 2017-02-23
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    Sure. $y^{\ast}$ is called a strong local minimum of $J$ over $C^1$ if there exists $\varepsilon > 0$ such that $J(y^{\ast}) \leqslant J(y)$ for all $y \in C^1[a,b]$ satisfying $\|y-y^{\ast}\|_{\infty} < \varepsilon$. For weak minimum, replace with $\|\cdot\|_1$. – 2017-02-23
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    So the $C^1$ norm case is easy enough, this is definitely a weak minimum (because once $|f'|<1$ everywhere, the integrand is positive, so for a function with a small positive $C^1$ norm, you will have a positive value of $J$). In the $C^0$ case, is it possible to have a $C^1$ function with arbitrarily small $C^0$ norm with a negative value of $J$? For instance, can you have a function which "almost always" has a derivative which is at least $2$ in absolute value, and which bounces back and forth between $\epsilon$ and $-\epsilon$? – 2017-02-23
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    Not necessarily, such a function could have wide swings within $(-\varepsilon,\varepsilon)$ giving large derivatives. Is this the idea? – 2017-02-23
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    Right, that's the point: if your function usually has a derivative of at least 2 in absolute value, but always swings back to the negative side, then you have large derivatives (and thus usually a large negative integrand to go into $J$) but you have small $C^0$ norm. – 2017-02-23

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Look at $f'_{\epsilon,\delta}$ given by the periodic extension of the piecewise linear interpolant of $(0,2),(\epsilon,2),(\epsilon+\delta,-2),(2\epsilon+\delta,-2),(2\epsilon+2\delta,2)$, where $\epsilon$ and $\delta$ are chosen so that $1/(2\epsilon+2\delta)$ is an integer. Then $f_{\epsilon,\delta}(x):=\int_0^x f'_{\epsilon,\delta}(y) dy$ will:

  • Be $C^1$.
  • Satisfy the boundary conditions (this is why we needed the periodicity structure).
  • Have $C^0$ norm on the order of $\epsilon$ provided $\delta$ is comparable to $\epsilon$ or smaller.
  • Have $J=-12+O(\delta)<0$ for small $\delta$ at fixed $\epsilon$. Here $-12$ is what you would get if the derivative simply bounced back and forth between $2$ and $-2$ without changing continuously in between.

This example (when the details are ironed out) will prove that $y=0$ is not a strong minimum of $J$. The idea of it is that a function with a small $C^0$ norm, even if it is $C^1$, is free to wiggle a great deal, as long as its wiggles always bring it back towards zero.