“Then obviously…” (a bound on strongly differential functions on an interval)

On the fourth page of their 1967 paper Local Behavior of Solutions of Quasilinear Parabolic Equations, Aaronson and Serrin comment: "Consider a strongly differentiable function $w$ of the real variable $x$, $0 < x < d$. Then obviously

$$|w(x)|^{2} \leq \frac{2}{d}\int_{0}^{d}|w|^{2}\ dx + 2d\int_{0}^{d}|w_{x}|^{2}\ dx$$

and..."

And obviously I don't get it, and am wondering what background I need to fill in before I can read this paper.

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2 Answers

This is I think just a very loose bound of the max norm by the norm of the function and the norm of the (strong) derivative. For any function $f$ defined over $(0,d)$ we have $$|f(x)| \leq \text{min}|f(t)| + \int_0^d |f'(t)| dt \leq \frac{1}{d}\int_0^d|f(t)|dt + \int_0^d |f'(t)| dt$$ All you need now is to replace $f(x)$ by $w^2(x)$. Note by AM-GM inequality $$\frac{1}{d}\int_0^d|w|^2dt + d\int_0^d|w_t|^2dt \geq 2\int_0^d|w\cdot w_t|dt=\int_0^d|(w^2(t)'|dt$$ and the inequality follows.

You probably don't need more back ground. You just need to treat each of the "obviously", "easy too see" etc statements as an exercise and try to solve it.

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Alternatively $$w(x)^2-w(y)^2=\int_y^x (w(u)^2)^{\prime}du = 2 \int_y^x w(u) w^\prime(u) du$$ Since $2ab \leq a^2/t +t b^2$ for any $t$, we have \begin{eqnarray*} w(x)^2&\leq& w(y)^2 + \frac{1}{t}\int_y^x w^2(u)du +t \int_{y}^{x} (w')^2(u)du \\ &\leq& w(y)^2 + \frac{1}{t}\int_0^d w^2(u)du +t \int_{0}^{d} (w')^2(u) du \end{eqnarray*} Now integrate this inequality with respect to $y$ between $0$ and $d$ $$d w(x)^2 \leq \int_0^d w(y)^2dy + \frac{d}{t}\int_0^d w^2(u)du +dt \int_{0}^{d} (w')^2(u)du$$ divide both sided by $d$ and choose $t=d$ to obtain $$w(x)^2 \leq \frac{2}{d}\int_0^d w(y)^2dy + d \int_{0}^{d} (w')^2(u)du.$$

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