Proof regarding p-norm of f

Question

Suppose that $f : [a,b] \to \mathbb R$ with $-\infty < a < b < \infty $ is continuous. Let $$M = \max_{a \leq x \leq b } |f(x)|.$$

I want to show that for any $\varepsilon > 0$, there is a constant $c > 0$ such that $c^{1/p}(M-\varepsilon)\leq \|f\|_{p}\leq (b-a)^{1/p}M.$

My working:

The $\|f\|_{p}\leq (b-a)^{1/p}M$ part is rather easy. I show that

$$ \|f\|_{p} =\bigg(\int_a^b |f(x)|^p\, dx\bigg)^{1/p} \leq \Bigg(\max_{a \leq x \leq b } |f(x)|^p\bigg(\int_a^b 1\, dx\bigg) \Bigg)^{1/p}=(b-a)^{1/p}M.$$

I'm having trouble proving $\|f\|_{p} \geq c^{1/p}(M-\varepsilon).$

Here is my partial working :

Let $1 \leq p,q \leq \infty$ with $\frac{1}{p} + \frac{1}{q}.$ Hölder's inequality for integrals states that $${\displaystyle \|fg\|_{1}\leq }{}\ \|f\|_{p}\|g\|_{q}.$$

Let $g(x) = 1.$ So, we have $\|f\|_{1} \leq (b-a)^{1/q}\|f\|_{p}.$

Now how do I show $\dfrac{\|f\|_{1}} {(b-a)^{1/q}} \geq c^{1/p}(M-\varepsilon) ?$

Answer 1

You cannot find a constant $c$ independent of $f$. If you allow it to depend on both $f$ and $\epsilon$ but not on $p$ then you can use the following argument:

There exists an interval $(c,d) \subset [a,b]$ such that $|f(x)| >M-\epsilon$ for $c < x<d$. We then get $\int_a^{b} |f(x)|^{p}dx \geq \int_c^{d} |f(x)|^{p}dx \geq (M-\epsilon)^{p} (d-c)$. So the required inequality holds with the constant $d-c$.

Answer 2

The statement is false, at least if $c$ is to be independent of $f$.

Fix $p\ge1$ and $\epsilon>0$ and suppose $$c^{1/p}(\|f\|_\infty-\epsilon)\le\|f\|_p$$ holds for some constant $c$.

Let $f_n$ be a triangle function with base on $[-1/n^{p+1},1/n^{p+1}]$ and height $n$, and zero otherwise. Then $\|f_n\|_\infty=n$ and $$\|f_n\|_p^p=2\int_0^{1/n^{p+1}}(n-n^{p+1}x)^p\,dx=\frac{2}{p+1}(1-(1-\tfrac{1}{n})^{p+1})\to0\ as\ n\to\infty$$

Then $c^{1/p}(n-\epsilon)\le \|f_n\|_p$ cannot hold as $n\to\infty$.