Let $f \in C^1[0, 1]$. Then show that the norms $$|f| =\int_0^1|f(t)|\, dt+\max_{t \in [0, 1]}|f'(t)|$$ and $$\|f\|= \max_{t \in [0, 1]}|f(t)|+\max_{t \in [0, 1]}|f'(t)|$$ are equivalent.
I have shown that $|f| \leq \|f\|$. How to show that $\|f\| \leq K |f|$ for some constant K?
On
Let $f$ be a function in the given space. It reaches the minimal and the maximal values, $m$ and $M$ say, at two points $a,b$ say inside $I=[0,1]$. Then the (only) term we have to control using $\int_I |f(x)|\; dx$ and $\|f'\|_\infty$ is $\|f\|_\infty=\max(|m|,|M|)$.
In the case $m,M$ have the same sign, we may and do assume (after possibly passing from $f$ to $-f$) $0\le m\le M$. Then $$ M =m+(M-m)\le \int_I|f(x)|\; dx+|f'(\xi)|\,|b-a|\le |f|\ . $$ In the remained case $m\le 0\le M$ we find an intermediate $c$ between $a$, $b$ with $f(c)=0$. So $$ \begin{aligned} |M| &= (M-0) = |f'(\xi_1)|\,|b-c|\le \|f'\|_\infty \le |f|\ , \\ |m| &= (0-m) = |f'(\xi_2)|\,|c-a|\le \|f'\|_\infty \le |f|\ . \end{aligned} $$ Summing up: $$ \|f\| =\|f\|_\infty + \|f'\|_\infty \le |f| + \|f'\|_\infty \le |f|+|f|= 2|f|\ . $$
Suppose that$$\lvert f\rvert\leqslant1.\tag1$$Then$$\displaystyle\max_{t\in[0,1]}\bigl\lvert f'(t)\bigr\rvert\leqslant1.\tag2$$Suppose that $\displaystyle\max_{t\in[0,1]}\bigl\lvert f(t)\bigr\rvert>2$. Then, by $(2)$ and by the Mean Value Theorem, $\displaystyle\min_{t\in[0,1]}\bigl\lvert f(t)\bigr\rvert>1$ and so $\displaystyle\int_0^1\bigl\lvert f(t)\bigr\rvert\,\mathrm dt>1$. This is impossible, since we are assuming $(1)$. This proves that $\lvert f\rvert\leqslant1\implies\lVert f\rVert\leqslant2$. Can you take it from here?