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

### 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.

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 : $$\|f\|_2 \leq (1 + \max(1+\|a\|_\infty, 1+\|b\|_\infty)C) \|g\|_\infty$$

• You are very right, I edited the question. Dec 21, 2021 at 14:00

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 $$\|x\|_2\le c\|Tx\|_\infty, \quad\text{for all x\in C^2[0,1]}.$$

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