Krdytkyu

I have a relatively simple analysis question that for some reason has me stuck.

Suppose that $f:mathbb Rtomathbb R$ satisfies the following conditions.

• 
  $f$ is differentiable


• 
  $|f|_2$ and $|f'|_2$ are finite


• 
  $f$ is compactly supported.

Then, a simple argument shows that $fleqsqrt{2|f|_2|f'|_2}$ pointwise:
$$|f(x)|^2=left|int_{-infty}^x(f(x)^2)'~dxright|leq2int_a^x |f(x)||f'(x)|~dxleq2|f|_2|f'|_2.tag{1}$$

Remark. Intuitively, this kind of makes sense: If $f$ has a small $L^2$-norm, then it can only take very large values on sets of small Lebesgue measure. But then since $f$ is continuous it probably has to increase/decrease at a large rate next to its very large values, which is "ruled out" by $f'$ having small $L^2$-norm.

Question. Does a similar result hold when we consider functions on a bounded interval $f:[a,b]tomathbb R$?

Replicating the same argument here yields the bound
$$|f(x)^2-f(a)^2|=left|int_{a}^x(f(x)^2)'~dxright|leq 2|f|_2|f'|_2,$$
but then this is not quite the same as my first example: if we don't know how large $f(a)$ is then we don't have a pointwise bound of $f$. However, I don't see why the intuition I mention in the remak above doesn't apply here.

In short, is my intuition misguided, or am I generalizing the procedure in equation $(1)$ incorrectly?

If $f$ has a small $L^2$-norm, then it can only take very large values on sets of small Lebesgue measure

It depends on what you mean by 'very large'. If you remove the restriction of compact support, then you also consider the constant functions $f(x)=c$ for all $xin [a,b]$. If we increase the constant $c$, these can take arbitrarily large values over the whole $[a,b]$ even if the derivative is $0$ everywhere for any choice of $c$.

In particular, in the RHS we have $$2|f|_2|f'|_2=0$$ which means that the correct term in the LHS is indeed $|f(x)^2-f(a)^2|$ and not $|f(x)|^2$ by itself or anything 'similar'. Your inequality is sharp.