How prove this inequaliy $\sup_{x \in [a, b]} |f(x)| \leq \frac{1}{b-a} \int_{a}^{b} |f(t)| dt + \int_{a}^{b} |f'(t)| dt $

Question

let $f \in C^1([a, b])$ with $a, b \in \mathbb{R}, a < b$
show that $$\sup_{x \in [a, b]} |f(x)| \leq \dfrac{1}{b-a} \int_{a}^{b} |f(t)| dt + \int_{a}^{b} |f'(t)| dt$$ I've tried to use the fundamental theorem of calculus and the mean value theorem for integrals, but this can't usefull. $$\dfrac{1}{b-a} \int_{a}^{b} |f(t)| dt + \int_{a}^{b} |f'(t)| dt \; \geq \; | \dfrac{1}{b-a} \int_{a}^{b} f(t) dt + \int_{a}^{b} f'(t) dt| \\ = | \dfrac{1}{b-a} \int_{a}^{b} f(t) dt + f(b) - f(a)| \; \geq \; |f(\xi) + f(b) - f(a)|$$ for $\xi \in (a, b)$.
How would I proceed? Any ideas?
Thank you very much!

Answer 1

For any $x,y \in [a, b]$,

\begin{equation} \begin{split} |f(x)| &\le |f(x) - f(y)| + |f(y)| \\ &= \left| \int_x^y f'(s)\, ds\right| + |f(y)| \\ &\le \int_a^b |f'(s)|\, ds + |f(y)| \end{split} \end{equation}

Now integrate this inequality with respect to $y$.

$$(b-a) |f(x)| \le (b-a) \int_a^b |f'(s)|\, ds + \int_a^b |f(y)|\, dy . $$

Answer 2

We have $\min_{[a,b]}|f| = |f(c)|$ for some $c\in [a,b].$ So for any $x\in [a,b],$

$$f(x) = f(c)+\int_c^x f'(t)\,dt \implies |f(x)| \le |f(c)| + \int_a^b |f'(t)|\,dt $$ $$= \frac{1}{b-a}\int_a^b|f(c)|\,dt + \int_a^b |f'(t)|\,dt \le \frac{1}{b-a}\int_a^b|f(t)|\,dt + \int_a^b |f'(t)|\,dt.$$