An elementary proof of Sobolev's Inequality in one dimension

Question

I was looking for an elementary proof for Sobolev's Inequality in one dimension and I found the proof in this paper, it's the proof of Sobolev imbedding. I would like an explanation about some steps of the proof.

There is a moment when the author of the paper asserts

$|f(y)| + |f(x) - f(y)| \leq \int_0^1 |f(t)|dt + |f(x) - f(y)| \leq \int_0^1 |f| \cdot 1 + |f'|_{L^2} \cdot 1$ $ \leq |f|_{L^2} + |f'|_{L^2} << 2 \left( |f|^2 + |f'|^2 \right)^{\frac{1}{2}},$

just a notice that $|f| := \left( \int_a^b |f(t)|^2 dt \right)^{\frac{1}{2}}$

I didn't understand in the first inequality why the author can asserts that $|f(y)| \leq \int_0^1 |f(t)| dt$. For this inequality, I think that we can assume without loss of generality that $f(0) = 0$, because we can translate $f$ so that $f(0) = 0$, therefore we would have

$|f(y)| = |f(y) - f(0)| = | \int_0^y f(t) dt | \leq | \int_0^1 f(t) dt | \leq \int_0^1 |f(t)| dt$

I didn't understand too why $\int_0^1 |f(t)|dt \leq \int_0^1 |f| \cdot 1$.

$\int_0^1 |f| \cdot 1 \leq |f|_{L^2}$

I don't sure, but it seems like that here is used Cauchy-Schwarz inequality.

$|f|_{L^2} + |f'|_{L^2} << 2 \left( |f|^2 + |f'|^2 \right)^{\frac{1}{2}}$

and I don't have idea why he can asserts this inequality.

$\textbf{EDIT1:}$

I forgot to ask why this proof allows to conclude that $\max |f(x)|^2 \leq C \int_0^1 \left( |f|^2 + |f'|^2 \right)$ for some constant $C$.

$\textbf{EDIT2:}$

I correct the EDIT1.

Thanks in advance!

Answer 1

I will use $\|f\|_X$ for function norms and $|f(x)|$ for absolute values.

You may have missed that the author states that $y$ was such that $|f(y)|= \min_x |f(x)| ≤ |f(z)|$ for every $z\in[0,1]$. Integrating in $z$,

$$ |f(y)| = \min_x |f(x)|\int_0^1 dz ≤ \int_0^1|f(z)| \, dz$$ You are correct for the next line, Cauchy Schwartz is used.

I will just write out the proof here,

\begin{align} |f(x)| &\overset{\text{triangle ineq}}≤ |f(y)| + |f(x)-f(y)| \\ &\overset{\text{definition of $y$}}≤ \int_0^1 |f(z)|\,dz + |f(x)-f(y)| \\ &\overset{\text{C-S ineq}}≤ \|f\|_{L^2}\int_0^11^2dz + \|f'\|_{L^2}|y-x|^{1/2} \\ &≤ \|f\|_{L^2} + \|f'\|_{L^2} \\ &≤ C(\|f\|_{L^2}^2 +\|f'\|_{L^2}^2)^{1/2} \end{align}

Where this last line is from the easy inequality $2ab≤a^2+b^2$ so that $$(a+b)^2 ≤ a^2 + b^2 + 2ab ≤ 2a^2+2b^2 $$ Take a square root to conclude. This line also follows from the fact that all norms on $\mathbb R^n$ are equivalent.

It should be noted that OP states a wrong inequality $\max_x |f(x)|^2 ≤ C(\|f\|_{L^2}^2 +\|f'\|_{L^2}^2)^{1/2}$ as can be seen from this line of reasoning.

Answer 2

Looking at the link, I actually don't think the first chain of inequalities is very clearly presented. Here's how I'd say the same thing:

Assume first that $f\in C^\infty$. Now for every $t$ it's clear that $$|f(x)|\le|f(t)|+|f(x)|+|f(x)-f(t)|.$$

Now since $|f'|=|f'|\cdot 1$, Cauchy-Schwarz shows that $$|f(x)-f(t)|=\left|\int_x^tf'\right|\le\int_x^t|f'|\le\int_0^1|f'| \le\left(\int_0^1|f'|^2\right)^{1/2}.$$So we have $$|f(x)|\le|f(t)|+\left(\int_0^1|f'|^2\right)^{1/2}.$$

Now take the integral from $0$ to $1$ of both sides of that last inequality, with respect to $t$. Since $f(x)$ and $\left(\int_0^1|f'|^2\right)^{1/2}$ are independent of $t$ (and of course $\int_0^1c=c$) we obtain $$|f(x)|\le\int_0^1|f(t)|\,dt+\left(\int_0^1|f'|^2\right)^{1/2}.$$Another application of Cauchy-Schwarz shows that $$|f(x)|\le\left(\int_0^1|f|^2\right)^{1/2}+\left(\int_0^1|f'|^2\right)^{1/2}.$$

Now if $|f(x)|\le c$ for every $x$ then $\max_x|f(x)|\le c$. So we've shown that $$\max_x|f(x)|\le \left(\int_0^1|f|^2\right)^{1/2}+\left(\int_0^1|f'|^2\right)^{1/2}.$$

That's the inequality we want, for $f\in C^\infty$. This automatically extends to $f\in H^1$, because of the following fact:

Suppose $X$ and $Y$ are normed vector spaces, $D$ is a dense subspace of $X$, and $T:D\to Y$ is a linear map with $||Tx||\le||x||$ for every $x\in D$. Then $T$ extends to a linear map $T:X\to Y$ with $||Tx||\le||x||$ for every $x\in X$.

(Apply that with $X=H^1$, $D=C^\infty$, and $Y=C([0,1])$, where $X$ has the Sobolev norm, $Y$ has the sup norm, and $Tf=f$.)