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Let $X=C([0,1])$ with the $\sup$ norm and let $T: X \to X$ given by $$ Tf(t) = f(t) + \int_0^t f(s) \,\mathrm{d}s. $$ I already showed that $T$ is a bounded linear operator with $\|T\|=2$. The problem asks me to show that $T$ is injective.

What I tried was to show $\ker T = \{0\}$, which means the equation $Tf(t)=0$, or \begin{equation} \tag{1} f(t) + \int_0^t f(s) \,\mathrm{d}s = 0 \end{equation} for all $t \in [0,1]$ has a unique solution $f \equiv 0$. Differentiating both sides with respect to $t$ gives the ODE: $$ f'(t) + f(t) = 0, $$ which has a solution $f(t)=Ce^{-t}$, where $C$ is a constant. Since equation $(1)$ gives $f(0)=0$, we must have $C=0$, and equation $Tf(t)=0$ has a solution $f \equiv 0$.

However, I'm not sure how to show $f = 0$ is the unique solution to $(1)$ (or maybe it must be from the general solution to the ODE).

I wonder if differentiation can be done in this case because $f$ is only continuous, not a $C^1$ function.

Is there other way to solve this problem like assuming $Tf = Tg$ and showing $f=g$ (I was stuck on this)?

Does it have something to do with the Hahn-Banach Theorem, (I don't know the space and the operator)?

Thank you.

EllaW
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2 Answers2

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One important point is that (1) directly implies $$ f(t) = -\int_0^t f(s) \, \mathrm d s,$$ i.e., $f$ is continuously differentiable, since the right-hand side is $C^1$ (fundamental theorem of calculus).

gerw
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It can be shown that the operator $T$ is invertible and verified directly that the inverse is given by the formula $$(T^{-1}f)(t)=f(t)-e^{t}\int\limits_0^te^{-s}f(s)\,ds$$ see 1 see 2. In particular the operator $T$ is injective

Ryszard Szwarc
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  • I really like your answer. I thought about Banach fixed-point theorem before, but the integral operator wasn't a contraction on $[0,1]$. You solved it. – EllaW Aug 01 '23 at 02:32
  • I have modified the answer as the operator itself is invertible – Ryszard Szwarc Aug 01 '23 at 03:45