Cfrgtkky

Let $f$ be a differentiable function. Denote a bilinear form by $$b(f,f) = int_{0}^{1} bigg( frac{d f(x)}{dx} bigg)^{2} dx.$$ Given $f(0) = 0,$ we want to show that $$a cdot b(f,f) geq ||f||_{infty}^{2},$$ where $||f||_{infty} = max_{0 leq x leq 1} |f(x)|.$ Note that $a$ is a constant independent of $f$.

In attempting the proof (which doesn't get far), I express $$||f||_{infty}^{2} = max_{0 leq x leq 1} |f(x)| cdot |f(x)|.$$ However, I am not sure of a bound for this quantity that will lead naturally to the desired proof. Also, this might not be the first step in the correct proof.

Note: By Holder's inequality we have $$bigg( int_{0}^{1} frac{d f (x)}{dx} dx bigg)^{2} leq int_{0}^{1} bigg( frac{d f(x)}{dx} bigg)^{2} dx.$$ The left hand side equals $(f(1) - f(0))^{2},$ which simplifies to $(f(1))^{2}$ since $f(0) = 0.$

Observe
begin{align}
|f(x)| = left|int^x_0 f'(t) dtright| leq int^1_0|f'(t)| dt leq left(int^1_0|f'(t)|^2 dt right)^{1/2}.
end{align}
Since this holds for all $x in [0, 1]$ then you are done.