estimate the max of a function using p-norm of the function and its derivative Prove the following interesting, elementary but tricky problem :
Let f be a smooth function defined on the interval $[a,b]$ and $p\in[1,\infty]$. Then there exists a constant $C_p$ such that for any $\mu\gt0$,$\max|f|\le C_p(|f(a)|+\mu^{1/p-1}||f'||_p+\mu^{1/p}||f||_p)$.
There is a hint for this exercise: using fundamental theorem of calculus and Holder inequality.
I started this problem by writing f(t) as f(a)+$\int_a^tf'(x)dx$. Then I can use Holder to get the $p$-norm of the derivative $f'$. However, I don't know how to relate this expression with the $p$-norm of f and $\mu$.
 A: You could use integration by parts to get 
$
\int_{a}^{t}f(x)^{p-1}f'(x)dx=f(t)^p-f(a)^p -(p-1)\int_{a}^{t}f(x)^{p-1}f'(x)dx
$,
so $f(t)^p=f(a)^p+p\int_{a}^{t}f(x)^{p-1}f'(x)dx$, then use the same trick as in the former answer. I believe it is true for all $1\leq p\leq \infty$.
A: We start with the cases $1<p<\infty$. From here, $p'$ is the conjugate exponent of $p$. Assume that $f\ge 0$; fix $\mu>0$ and write (Fundamental Theorem of Calculus) $$f(x)^p=f(a)^p+\int_a^x[f(s)^p]'ds\tag{1}$$
$(1)$ can be rewritten as $$f(x)^p=f(a)^p+p\int_a^x\mu^{1/p'}f(t)^{p-1}\frac{f'(t)}{\mu^{1/p'}}dt\tag{2}$$
Now we use Holder inequality to conclude that $$f(x)^p\le f(a)^p+p\mu^{-1/p'}\|f'\|_p\mu^{1/p'}\|f\|_p^{p/p'}\tag{3}$$
We use the inequality $ab\le \frac{a^p}{p}+\frac{b^{p'}}{p'}$, $a,b\ge 0$, to conclude from $(3)$ that $$f(x)^p\le f(a)^p+p\left(\frac{\mu^{-p/p'}\|f'\|_p^p}{p}+\frac{\mu\|f\|_p^p}{p'}\right)\tag{4}$$
We have from $(4)$ that there is $C_p>0$ such that $$f(x)\le C_p(f(a)+\mu^{-1/p'}\|f'\|_p+\mu^{1/p}\|f\|_p)\tag{5}$$
Therefore, if $f\ge 0$, we conclude from $(5)$ the desired inequality. I will leave to you the case where $f$ can change sign.
With respect to the cases $p=1,\infty$, they are no true. Indeed, assume for example that there is a constant $C_1>0$ depending only in $p=1$ such that $$\|f\|_\infty\le C_1(|f(a)|+\mu\|f\|_1)\tag{6}$$
Letting $\mu\to 0$ in $(6)$ we get $$\|f\|_\infty\le C_1|f(a)|,\ \forall f\tag{7}$$
Well, $(7)$ is readily seen to be not true. An analogous argument works for the case $p=\infty$.
