Let $\;f \in W^{1,2}_{loc}((a,b);\mathbb R^n)\;$ where $\; -\infty \le a \lt b \le \infty\;$ and the choice of $\;a,b\;$ depends on $\;f\;$.
My professor told me that if I could show that $\; \nexists \;x \in (a,b)\;$ such that $\; \vert f(x) \vert = \infty\;$ then $\; \exists \;C \gt 0\;$ such that $\; {\vert \vert f \vert \vert}_{L^{\infty}((a,b);\mathbb R^n)}\;\le C\;$.
Why is that true? All I understand is that if there isn't $\;x \in (a,b)\;$ such that $\; \vert f(x) \vert = \infty\;$ then $\; \vert f(x) \vert \le C\;$ for some constant $\;C \gt 0\;$. But how does this connect to the $\; L^{\infty}\;$ bound?
I 'm having a really hard time getting my head around this. Any help would be valuable!
Thanks in advance!
This is not true. There is a big difference between $W^{1,2}(a,b)$ and $W^{1,2}_{\text{loc}}(a,b)$. Any function $f\in C^1((a,b))$ will belong to $W^{1,2}_{\text{loc}}(a,b)$, in particular $f(x)=\frac1{x}$ belongs to $W^{1,2}_{\text{loc}}(0,1)$ but it is not bounded. However if $f\in W^{1,2}(a,b)$ then (because a representative is absolutely continuous) you can apply the fundamental theorem of calculus to get $$f(x)-f(y)=\int_{y}^xf'(t)\,dt$$ and so $$\vert f(x)\vert^2\le 2\vert f(y)\vert^2+2\left(\int_{a}^b\vert f'(t)\vert \,dt\right)^{2}.$$ Integrating in $y$ gives $$(b-a)\vert f(x)\vert^2\le 2\int_{a}^b\vert f(y)\vert^2dy+2(b-a)\left(\int_{a}^b\vert f'(t)\vert \,dt\right)^{2}.$$ By Holder's inequality $$(b-a)\vert f(x)\vert^2\le 2\int_{a}^b\vert f(y)\vert^2dy+2(b-a)^2\int_{a}^b\vert f'(t)\vert^2 \,dt.$$ Hence $$\Vert f\Vert_{L^\infty}=\sup_x\vert f(x)\vert^2\le \frac2{b-a}\int_{a}^b\vert f(y)\vert^2dy+2(b-a)\int_{a}^b\vert f'(t)\vert^2 \,dt.$$