Derivative w.r.t. the Lp norm

[foxjwill]

Homework Statement

Let $$A:L^2(0,\infty)\to L^2(0,\infty)$$ be given by $$f(x)\mapsto f(x+1)$$. What is the derivative $$A'$$, if it exists, of $$A$$? That is, we want a function $$A':L^2(0,\infty)\to L^2(0,\infty)$$ such that

$$\lim_{\|h\|\to0}\frac{\left\|A(f+h)-Af -hA'f\right\|}{\|h\|}=0.$$​

Homework Equations

The Attempt at a Solution

I was able to simplify the differentiability condition thingy to

$$\lim_{\|h\|\to0}\frac{\left\|h(x+1) -h(x)A'f\right\|}{\|h(x)|}=0.$$​

However, I'm not sure where to go from there. Any hints?

 

[mighty2000]

Hi there! It looks like you're on the right track with simplifying the differentiability condition. To find the derivative A', we can use the definition of the derivative as the limit of the difference quotient. So, let's start by expanding the numerator of the difference quotient:

\|A(f+h)-Af -hA'f\| = \|f(x+1+h)-f(x+1) -hA'f\|

Now, we can use the fact that A is given by f(x)→f(x+1) to simplify this further:

\|f(x+h)-f(x) -hA'f\| = \|f(x+h)-f(x) -hA'f\|

Next, we can use the definition of the norm to expand the norm of the difference:

= \sqrt{\int_{0}^{\infty}|f(x+h)-f(x) -hA'f|^2 dx}

= \sqrt{\int_{0}^{\infty}|f(x+h)-f(x)|^2 - 2h(f(x+h)-f(x))A'f + h^2|A'f|^2 dx}

Now, we can use the fact that f is in L^2(0,\infty) to simplify the first term inside the square root:

= \sqrt{\int_{0}^{\infty}|f(x+h)-f(x)|^2 dx + h^2\|A'f\|^2}

Next, we can use the triangle inequality to split up the first term inside the square root:

= \sqrt{\int_{0}^{\infty}|f(x+h)-f(x)|^2 dx} + h\|A'f\|

Finally, we can divide both sides by \|h\| and take the limit as \|h\|→0 to get:

\lim_{\|h\|\to0}\frac{\left\|A(f+h)-Af -hA'f\right\|}{\|h\|} = \lim_{\|h\|\to0}\frac{\sqrt{\int_{0}^{\infty}|f(x+h)-f(x)|^2 dx} + h\|A'f\|}{\|h\|}

Now, we can use the fact that f is in L^2(0,\infty) to show that