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Problem (Unsolvable, since the statement is false.)

Let $X = (C^2([0,1]),||\cdot||_2)$, where $||f||_2 = ||f''||_{\infty} + ||f'||_{\infty} + ||f||_{\infty}$.

Define $T:X\to C([0,1])$ by: $\ \ \ \ T(f) = f'' + af' + bf$

where $a,b \in C([0,1])$ such that $T$ is surjective.

Prove that $\exists c > 0$ such that $||f||_2 \le c ||T(f)||_{\infty}$

What I've gathered

I know that $X$ is complete and that $T$ is linear. $T$ is bounded as well, since for $||f||_2 = 1$:

$||T(f)||_\infty = ||f'' + af' + bf||_\infty \le ||f''||_\infty + A ||f'||_\infty + B ||f||_\infty \le (1+A + B)$.

where $A = ||a||_\infty, B=||b||_\infty$. So far, I know that $T$ is a bounded linear map, so I know that:

$\forall f \in X: ||T(f)||_\infty \le ||A||\cdot||f_2||$

So what I need to prove is that $\exists c > 0$ such that $\forall f \in X:$

$\frac{||T(f)||_\infty}{||A||} \le ||f||_2 \le c ||T(f)||_\infty$

i.e. that $||\cdot||_2$ and $||T(\cdot)||_\infty$ are equivalent norms, right?

I have no idea how to proceed from here..

Edit

Since I got a counterexample showing that above is not true in general, I show below the original question (taken from an exam), which I shortened, but maybe changed the question in doing so:

Original Problem

Consider the Banach space $C^2([0,1])$ with $||\cdot||_2$ as defined above. Let $a,b \in C([0,1])$ and assume that for every $g\in C([0,1])$, the differential equation $f''+ a f' +b f = g$ has a solution $f\in C^2([0,1])$. Prove that there exists $c > 0$, such that for every $g \in C([0,1])$, the above equation has a solution $f\in C^2([0,1])$, with $||f||_2 \le c ||g||$

Problem $\neq$ Original Problem

I realize now that the property $ \forall f: ||f||_2 \le ||T(f)||_\infty $ is not explicitly required. My bad.

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2 Answers 2

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This result is false, here is a counterexample :

Take $a= 0$ and $b=1$. We first show that $T$ is surjective.

Let $g\in C([0,1])$. Then : $$f(t) = \int_0^t \sin(t-s)g(s)\text ds$$ is in $C^2([0,1])$ and has $Tf = g$.

Then, if you take $f = \cos$, you get $T(f) = 0$ but $f\neq 0$, so there can be no constant $c>0$ such that : $$\|f\|_2 \leq c \|Tf\|_\infty$$

More generally, it seems that for any $a,b \in C([0,1])$, the theory of linear differential equations guarantees that $T$ has a non trivial kernel, so that this result is false in every case.

Edit : new question

To show that there is a constant $c >0$ such that for any $g\in C([0,1])$ there is a $f\in C^2([0,1])$ such that $T f =g$ and $\|f\|_2 \leq c\|g\|_\infty$, we rewrite the equation $f'' + af' + bf = g \tag{1}$ as a first order differential equation : let $$X = \begin{pmatrix} f' \\ f\end{pmatrix}, \quad A = \begin{pmatrix} -a & -b \\ 1 & 0\end{pmatrix} \quad \text{and}\quad B = \begin{pmatrix} g \\ 0 \end{pmatrix}$$ Then $(1)$ is equivalent to : $$X' = AX +B \tag 2$$

Let $U \in C^1([0,1],\mathcal M_2(\mathbb R))$ be then solution to the Cauchy problem : $$\left\{ \begin{array}{ll} U'= AU \\U(0) = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\end{array}\right.$$ It exists (and is unique) by the Cauchy-Lipschitz theorem. Then, we can check that : $$X(t) = U(t)\int_0^t U^{-1}(s) B(s)\text ds$$

is a solution of $(2)$. There is therefore $f\in C^2([0,1])$ such that : $$X = \begin{pmatrix} f' \\f\end{pmatrix}$$

Then, with $\| \cdot \|_1$ the $L^1$ norm on $\mathbb R^2$ and $|||\cdot|||$ the associated operator norm, we have : \begin{align} \| f \|_\infty + \|f' \|_\infty &= \sup_{t\in [0,1]} \|X(t)\|_{1} \\ &\leq C\| g\|_\infty \end{align} with $C = \sup_{t\in [0,1]} ||| U(t)||| \times \sup_{t\in [0,1]}|||U^{-1}(t) ||| < \infty$.

To conclude, we compute: $$\| f'' \|_\infty = \| g \| + \|a\|_\infty \|f'\|_\infty + \| b\|_\infty \|f\|_\infty$$ so : $$\|f\|_2 \leq (1 + \max(1+\|a\|_\infty, 1+\|b\|_\infty)C) \|g\|_\infty$$

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  • $\begingroup$ You are very right, I edited the question. $\endgroup$
    – JustANoob
    Dec 21, 2021 at 14:00
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This is true provided that $T$ is injective, i.e., $Tf=0\Longrightarrow f=0$.

In such case, since $T$ is surjective, and both $C[0,1]$ and $C^2[0,1]$ are Banach spaces, then $T$ is open, due to Open Mapping Theorem, and hence $T^{-1}$ is continuous and therefore bounded, and thus there exists a $c>0$, such that $$ \|T^{-1}y\|_2\le c\|y\|_\infty, \quad\text{for all $y\in C[0,1]$} $$ equivalently $$ \|x\|_2\le c\|Tx\|_\infty, \quad\text{for all $x\in C^2[0,1]$}. $$

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  • $\begingroup$ It turns out it is not injective. And my question was wrong, see edit. $\endgroup$
    – JustANoob
    Dec 21, 2021 at 14:01

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