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I know that function evaluation in $H^1(0,1)$ is continuous (see, e.g., Is the Delta distribution a continuous functional on $H^1(\mathbb R)$).

So, $\delta_x : H^1(0,1)\to\mathbb C$ is a continuous operator. Since it has finite-dimensional range, it is compact. In other words, $\delta_x$ is $D$-compact in $L^2(0,1)$, where $Df = f'$, $f\in H^1(0,1)$.

As $D$-compact operators have $D$-bound zero, for any $\varepsilon>0$, I should find an $a > 0$ such that $$ |f(x)|\,\le\,a\|f\|_2 + \varepsilon\|f'\|_2,\quad f\in H^1(0,1). $$ But I can't find such $a$ for given $\varepsilon$. Can anyone help me out?

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  • $\begingroup$ What is $H^1(0,1)$? $\endgroup$ – user56834 Mar 23 at 6:14
  • $\begingroup$ @user56834 It's the usual Sobolev space over $(0,1)$ based on $L^2(0,1)$. $\endgroup$ – amsmath Mar 23 at 6:20
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A somehow clumsy argument: By approximation assume $f$ is $C^1$. First assume $x\leq \frac 12$. If $|f(y)|>\frac 12 |f(x)$| for all $y\in [x, x+\frac 14 \epsilon^2]$, then $$\int_0^1f^2(y)dy\geq \int_x^{x+\frac 14 \epsilon^2}f(y)^2 dy\geq \frac 14\epsilon^2\frac 14 f(x)^2; $$ thus in this case $|f(x)|\leq 4\epsilon^{-1}\|f\|_2$. If for some $y\in [x, x+\frac 14 \epsilon^2]$ we have $|f(y)|\leq \frac 12 |f(x)|$, then $$ f(x)^2\leq 4[f(x)-f(y)]^2=4\Big(\int_x^y -f'(t) dt\Big)^2 \leq 4\int_x^y|f'(t)|^2dt \cdot |y-x| \leq \epsilon^2\int_0^1|f'(t)|^2dt, $$ i.e. $|f(x)|\leq \epsilon \|f'\|_2$.

If $x>\frac 12$ we can argue similarly by considering $y\in [x-\frac 14 \epsilon^2, x]$.

Overall we see that one can take $a=4 \epsilon^{-1}$.

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  • $\begingroup$ Thank you very much!!! :o) Actually, I don't see any point where you need that $f\in C^1$. $\endgroup$ – amsmath Mar 23 at 15:12
  • $\begingroup$ Before we proved this inequality, we don't know if we can define $f(x)$ (i.e. one tries to make sense of the value of an$L^2$ function at a particular point), as well as the use of fundamental theorem of calculus. So one starts with smooth functions then get the general case by approximation. $\endgroup$ – Yu Ding Mar 23 at 15:59
  • $\begingroup$ Yu, please note that $f\in H^1(0,1)$ implies that $f$ has an absolutely continuous representative. And for these you have the generalized (Lebesgue-) fundamental theorem of calculus. So, everything is nice and well settled. $\endgroup$ – amsmath Mar 23 at 16:42
  • $\begingroup$ Well, you have to quote that result, which in my view is not completely "trivial", because it (continuous representative) is not true for higher dimensions... $\endgroup$ – Yu Ding Mar 23 at 17:24
  • $\begingroup$ It is neither trivial that $C^1$ is dense in $H^1$. ;-) $\endgroup$ – amsmath Mar 23 at 17:43

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