Let $X$ be the Banach space $(C^1[0,1],\lVert\cdot\rVert)$, where
\begin{equation*}
\lVert f\rVert=|f(0)|+\max_{t\in[0,1]}|f'(t)|.
\end{equation*}
We denote $Y=(C^1[0,1],\lVert\cdot\rVert_I)$, where
\begin{equation*}
\lVert f\rVert_I=\int_{0}^1|f(t)|\,dt+\int_{0}^1|f'(t)|\,dt.
\end{equation*}
I want to prove that this two norms are equivalent. In order to do that, I'm trying to prove that $i:X\rightarrow Y $ is bicontinuous ($i$ and $i^{-1}$ are continuous). Since $i$ and $i^{-1}$ are linear, I only have to show that they are bounded.
To prove that $i$ is bounded, I've tried this:
\begin{equation*}
\lVert i(f)\rVert_I=\lVert f\rVert_I=\int_0^1|f(t)|\,dt+\int_{0}^1|f'(t)|\,dt\leq\int_0^1|f(t)|+\max_{t\in[0,1]}|f'(t)|
\end{equation*}
But I don't know how to follow from here. I've also got stuck proving that $i^{-1}$ is bounded.
How do you think I should follow the proof? Is there another easier or clever approach? Thanks.
Best Answer
The norms don't seem to be equivalent.
For one direction, let $f \in C^1[0,1]$ and recall that by the mean value theorem for any $x \in \langle 0,1\rangle$ there exists $\theta \in \langle 0,x\rangle$ such that $$|f(x)-f(0)| \le |x-0||f'(\theta)| = |x| |f'(\theta)| \le \max_{t \in [0,1]}|f'(t)|.$$ Therefore $$|f(x)| \le |f(x)-f(0)| + |f(0)| \le |f(0)| + \max_{t \in [0,1]}|f'(t)|.$$ Integrating this we get \begin{align*} \|f\|_I &= \int_0^1 |f(x)|\,dx + \int_0^1 |f'(x)|\,dx \\ &\le \int_0^1 (|f(0)| + \max_{t \in [0,1]}|f'(t)|)\,dx + \int_0^1 (\max_{t \in [0,1]}|f'(t)|)\,dx\\ &= |f(0)| + 2\max_{t \in [0,1]}|f'(t)|\\ &\le 2\|f\|. \end{align*}
The other inequality cannot be obtained. Consider $f_n(x) = x^n$ for $n \in\Bbb{N}$. Then $$\|f_n\| = |0^n| + \max_{t\in[0,1]} |nt^{n-1}| = n \xrightarrow{n\to\infty} +\infty$$ but $$\|f_n\|_I = \int_0^1 t^n \,dt + \int_0^1 nt^{n-1} \,dt = \frac1{1+n} + 1 \xrightarrow{n\to\infty} 1.$$