I am dealing with the Sobolev space $W^{m,2}[0,1]$, i.e. the space of functions on $[0,1]$ with absolutely continuous $m-1$st derivative and square integrable $m$th derivative. I am using the norm
$$||f||_{W} := \sqrt{\int_{0}^1 f^2(t) dt + \int_{0}^1 f^{(m)2}(t) dt },$$
but for my problem it is more convenient to work with the norm
$$||f||_{W,h} := \sqrt{\int_{0}^1 f^2(t) dt + h^{2m}\int_{0}^1 f^{(m)2}(t) dt },$$
where $h$ is a real number in $(0,1]$. My question is then how to extend the sobolev embedding theorem to this case. That is, if there exists a constant $c_m$ depending only on $m$ such that
$$ |f(x)| \leq c_m ||f||_{W}$$
for all $x\in[0,1]$, how can I show that there exists a constant $c_m$, possibly even the same constant, such that
$$|f(x)| \leq c_m h^{-1/2} ||f||_{W,h}. $$
I have been able to show that the above holds for $x \in [0,h]$ by defining $g(x) := f(xh)$ and applying the previous embedding but I am having trouble extending this result to all of $[0,1]$, which is the domain off interest.
Thank you in advance.
Here is a partial solution in the case $m=1$. If $f \in W^{1,2}$ then $f$, and consequently $f^2$, are AC on $[0,1]$ so that for $x,y \in [0,1]$ you have $$f(x)^2 = f(y)^2 + \int_y^x (f(t)^2)' \, dt = f(y)^2 + 2 \int_y^x f(t) f'(t) \, dt.$$
For any $\epsilon > 0$ you have $$|f(t) f'(t)| \le \frac{|f(t)|^2}{2 \epsilon} + \frac{\epsilon}2 |f'(t)|^2$$ so that you may integrate $dt$ to get $$f(x)^2 \le f(y)^2 + \frac 1{\epsilon} \|f\|_2^2 + \epsilon \|f'\|_2^2$$ and again $dy$ to get $$f(x)^2 \le \left(1 + \frac 1\epsilon \right) \|f\|_2^2 + \epsilon \|f'\|_2^2.$$
Take the square root to obtain $$|f(x)| \le \sqrt{1 + \frac 1\epsilon} \sqrt{ \|f\|_2 + \frac{\epsilon^2}{1 + \epsilon} \|f'\|_2^2}$$
For any $h \in (0,1]$ you can select $\epsilon \in (0,\frac{1 + \sqrt 5}{2})$ satisfying $h^2 =\dfrac{\epsilon^2}{1 + \epsilon}$. For this particular choice of $\epsilon$ you get $$|f(x)| \le \sqrt{1 + \frac 1\epsilon} \|f\|_{W,h}.$$
Clearly $h \le \epsilon$ and since $\epsilon < 3$ you get $$1 + \frac 1\epsilon = \frac{1 + \epsilon}{\epsilon} \le \frac{4}\epsilon \le \frac 4h$$ giving you $$ |f(x)| \le \frac{2}{\sqrt{h}} \|f\|_{W,h}.$$
Here $c_1 = 2$ is the correct constant.