Uniform convergence by using Fréchet derivative

Question

I was reading Theorem 1.5 at the page 187 from the book ''Semigroup of Linear operators and Application To Partial Differential Equations'' by ''A Pazy'' and I could not understand some step that is given beow.

Let $X$ be a Banach space and $f:[t_0,T]\times X\to X$ is continuously differentiable from $[t_0,T]\times X$ into $X$ for some $t_0\geq0.$ Let $u\in C([t_0,T];X)$ and $B(s)=(\partial /\partial x)f(s,x) $ for $s\in [t_0,T].$ Let $$w_1(s,h)=f(s,u(s+h))-f(s,u(s))-B(s)(u(s+h)-u(s))$$ and $$w_2(s,h)=f(s+h,u(s+h))-f(s,u(s+h))-(\partial/\partial s)f(s,u(s+h)).h.$$ In the Theorem, it says that $\frac{1}{h}w_i(s,h)\to 0$ uniformly on $[t_0,T]$ for $i=1,2.$ I can not understand the step $\frac{1}{h}w_i(s,h)\to 0$ uniformly on $[t_0,T]$. Please help me.

Answer 1

I figured I would write what I wrote down as some help, as it contains at least the result that you need to change some conditions.

The statement is not true as it is given. The condition of $u$ simply being continuous is not enough. As a counter example choose $[t_0,T]=[-1,1]$, $X=\mathbb R$, $f(s,x)=x^2$, $u(s)=s^{1/3}$. Then if you evaluate $\frac1h w_1(0,h)$ you get $$\frac{h^{2/3}}h-0\cdot \frac{h^{2/3}}h=\frac1{h^{1/3}}$$ which does not even converge, let alone converge uniformly to $0$.

The statement should be true if you assume $u$ is continuously differentiable. In this case it looks very much like an application of the differential quotient of a continuously differentiable function always converging uniformly on compacta to the derivative.

Let $f: [a,b]\to V$ be continuously differentiable with $V$ a complete normed space. Then the differential quotient $$\frac{f(t+h)-f(t)}{h}$$ converges uniformly to $f'(t)$ as $h\to0$.

To prove this statement:

The function $f'$ is a continuous function on a compact set, and thus uniformly continuous. So let $\epsilon>0$, there exists a $\delta>0$ so that if $|t-x|<\delta$ one has $|f'(x)-f'(t)|<\epsilon$. Let $|h|<\delta$, together with: $$\left|\frac{f(t+h)-f(t)}h-f'(t)\right|=\left|\frac1h\int_{t}^{t+h} (f'(x)-f'(t))\ dx\right|≤\frac1h \ \epsilon h$$ One gets uniform convergence of the differential quotient.

Look at a term in $\frac1hw_1$:

$$\frac{f(s,u(s+h))-f(s,u(s))}h=\frac{f(s,u(s+h))-f(s,u(s))}{u(s+h)-u(s)}\frac{u(s+h)-u(s)}h$$ Plug into the larger expression: $$\frac1h w_1(s,h)=\left(\frac{f(s,u(s+h))-f(s,u(s))}{u(s+h)-u(s)}-B(s,u(s))\right) \frac{u(s+h)-u(s)}h$$ Now the right term converges uniformly to $u'(s)$, but you cannot directly use that the left term converges uniformly to zero, rather you could use something like $$\frac{f(t,u(s+h))-f(t,u(s))}{u(s+h)-u(s)}-B(t,u(s))$$ converges uniformly to $0$ in $s$ for all $t\in[t_0,T]$.