most recent 30 from http://mathoverflow.net 2013-05-26T04:59:01Z http://mathoverflow.net/feeds/user/11512 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98266/an-inequality-with-ell-p-norm/98273#98273 Elias Pipping 2012-05-29T12:49:07Z 2012-05-29T13:15:22Z

<p>The way it currently stands, $L$ is the intersection of the "positive" part of the unit sphere w.r.t. the $\ell^1$-norm and an affine hyperspace. If $a_i$ and $b$ are allowed to be arbitrary, $L$ will in general not be non-empty; it could also contain just a single element.</p>

http://mathoverflow.net/questions/63462/trace-theorem-for-ck-1-domains/63513#63513 Elias Pipping 2011-04-30T08:22:35Z 2012-04-25T13:52:48Z

<p>The aforementioned article is now available from <a href="http://dx.doi.org/10.1155/2007/471535" rel="nofollow">here</a>.</p> <p>Abstract:</p> <blockquote> We prove that the well-known trace theorem for weighted Sobolev spaces holds true under minimal regularity assumptions on the domain. Using this result, we prove the existence of a bounded linear right inverse of the trace operator for Sobolev-Slobodeckij spaces $W_p^s(\Omega)$ when $s-1/p$ is an integer. </blockquote>

http://mathoverflow.net/questions/66361/an-inequality-on-concave-functions/66646#66646 Elias Pipping 2011-06-01T10:35:15Z 2011-06-01T10:35:15Z

<p>We want $$f(x)f(stx)\le f(sx)f(tx)$$ for concave non-decreasing functions with $f(0) = 0$. Since we require this for every $x$ and $f$ we can assume $x = 1$, because our claim is invariant w.r.t scaling in the argument. That means $$f(1)f(st)\le f(s)f(t)$$ Now we're looking at a function $f \colon [0,1] \to \mathbb R^+$ that is continuous, thus bounded; without loss of generality again $f(1) = 1$ because our claim is invariant w.r.t scaling again (let's leave the case $f(1) = 0$ aside for a second). So we're looking at $$f(st)\le f(s)f(t)$$ for a concave non-decreasing function $f \colon [0,1] \to [0,1]$ with $f(1) = 1$.</p> <p>Maybe this gets us any further?</p>

http://mathoverflow.net/questions/66361/an-inequality-on-concave-functions/66387#66387 Elias Pipping 2011-05-29T18:37:05Z 2011-05-29T18:37:05Z

<p>That $\log \circ f$ is concave follows from concavity of $\log$ and $f$ because $f$ is <strong>non-decreasing</strong>. I do not see how you could put positivity to use here.</p>

http://mathoverflow.net/questions/66361/an-inequality-on-concave-functions/66646#66646 Elias Pipping 2011-06-01T10:36:30Z 2011-06-01T10:36:30Z

The case $f(1) = 0$ is actually irrelevant because then we'd have $f = 0$.

http://mathoverflow.net/questions/66361/an-inequality-on-concave-functions/66387#66387 Elias Pipping 2011-05-30T09:47:24Z 2011-05-30T09:47:24Z

I meant that positivity of $f$ does not help in determining whether or not $log \circ f$ is concave. My answer should've been a reply to wmmiao's answer of course but I lack the permission to reply.