# $\forall\varepsilon > 0,\exists\ a >0 : |f(x)|\,\le\, a\|f\|_2 + \varepsilon\|f'\|_2$ for $f\in H^1(0,1)$

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?

• What is $H^1(0,1)$? – user56834 Mar 23 at 6:14
• @user56834 It's the usual Sobolev space over $(0,1)$ based on $L^2(0,1)$. – amsmath Mar 23 at 6:20

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 $$a=4 \epsilon^{-1}$$.
• Thank you very much!!! :o) Actually, I don't see any point where you need that $f\in C^1$. – amsmath Mar 23 at 15:12
• 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. – Yu Ding Mar 23 at 15:59
• 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. – amsmath Mar 23 at 16:42
• It is neither trivial that $C^1$ is dense in $H^1$. ;-) – amsmath Mar 23 at 17:43