Show that $u \in H^1(\omega)$ implies $|u(0)|\leq\sqrt{2}\|u\|_{H^1}$

Question

I'm trying to show that $|v(0)| \leq \sqrt{2}\|v\|_{H^1}$ we have:

$$ |v(x)| = \left|\int_{0}^{x} v'(t) dt\right| \leq \int_{0}^{x} |v'(t)| dt \leq x^{\frac 1 2} \|v\|_{H^1} $$ if you take $x = 0$, it doesn't go anywhere.

Answer 1

You began by misapplying the fundamental theorem of calculus. You should have $$|v(x) - v(0)| = \left| \int_0^x v'(t) \, dt\right|\le \int_0^1 |v'(t)| \, dt$$ so that $$|v(0)| \le |v(x)| + \int_0^1 |v'(t)| \, dt$$ for any $x \in (0,1)$. Now integrate from $0$ to $1$ and apply Holder's inequality: $$|v(0)| = \int_0^1 |v(0)| \, dx \le \int_0^1 |v(x)| \, dx + \int_0^1 |v'(t)| \, dt = \int_0^1 |v(t)| + |v'(t)| \, dt$$ $$|v(0)| \le \left( \int_0^1 (|v(t)| + |v'(t)|)^2 \, dt \right)^{1/2}.$$ The elementary inequality $(a+b)^2 \le 2(a^2 + b^2)$ gives you $$|v(0)| \le \left(2 \int_0^1 |v(t)|^2 + |v'(t)|^2 \, dt \right)^{1/2}$$ which is what you needed.

Answer 2

1. In your inequality, taking $x=0$ gives the best bound in the world which is too good (it ‘proves’ that $u(0)=0$ so of course it’s bounded by the $H^1$ norm.)
2. Hint- you used the FTC wrong. It’s not $\int f’ = f(x)$, it’s $\int f’=f(x)-f(0)$.