$I-T : L^p(\mathbb R) \rightarrow L^p(\mathbb R)$ is not surjective Let $p > 1$. Define $T : L^p(\mathbb R) \rightarrow L^p(\mathbb R)$ by $Tf(x) := \int_0^1 f(x+y)\,dy$.
Prove that $I-T$ is not surjective, where $I : L^p(\mathbb R) \rightarrow L^p(\mathbb R)$ is the identity map.
Using Jensen's inequality, one can show that $\|Tf\|_p \leq \|f\|_p$, so in particular, $T$ is well-defined, bounded linear operator with $\|T\| \leq 1$.
Beyond this, I don't have much idea how to proceed. I would appreciate any hint or reference.
 A: What follows gives the full answer for $1<p\le 2$ and partial answer for $p>2.$
First we are going to show that the operator $T-I$ is not invertible for $p>1.$
On the way will make use of the inequality
$$ 1-e^{-t}\le t,\quad 0\le t\le 1\qquad (*)$$
For $0<a<1$ let $$f_a(x)=\begin{cases}e^{-ax} & x\ge 0\\
0 & x<0
\end{cases}
$$ Then
$$(Tf_a)(x)=
\begin{cases}
a^{-1}[1-e^{-a}]e^{-ax} & x\ge 0
\\  a^{-1}[1-e^{a(x+1)}] & -1<x<0
\\
0 & x\le -1
\end{cases}
$$
Hence
$$[(T-I)f_a](x)
=a^{-1}[1-e^{-a}-a]f_a+g_a(x)$$
where
$$g_a(x)=  \begin{cases} a^{-1}[1-e^{-a(x+1)}] & -1<x<0\\
0 & x\le -1,\ x>0
\end{cases}
$$
By $(*)$ we get
$$g_a(x)\le \begin{cases} x+1 & -1<x<0\\
0 &  x\le -1,\ x>0
\end{cases}$$
Hence $\|g_a\|_p\le 1$ and
$$\|(T-I)f_a\|_p\le a^{-1}[e^{-a}+a-1]\,\|f_a\|_p+\|g_a\|_p \le a^{-1}[e^{-a}+a-1]\,\|f_a\|_p+1 $$
We have
$\|f_a\|_p=(ap)^{-1/p}.$ Thus
$${\|(T-I)f_a\|_p\over \|f_a\|_p}\le  a^{-1}[e^{-a}+a-1]+(ap)^{1/p}$$
The right hand side tends to $0$ when $a\to 0^+.$ Therefore the operator $T-I$ is not invertible.
Next we will  show that $I-T$ is injective for $1\le p\le 2$ by   applying the Fourier transform $f\mapsto\widehat{f}.$ This mapping  is an isometry on $L^2(\mathbb{R}),$ and is a bounded injective operator from $L^p$ to $L^q$ for $q=p/(p-1).$
Assume  $f=Tf.$  Applying the Fourier transform gives
$$\widehat{f}(\xi)=\widehat{Tf}(\xi)= \left (\int\limits_0^1e^{i\xi t}\,dt\right )\hat{f}(\xi)={e^{i\xi}-1\over i\xi}\widehat{f}(\xi)\quad (*)$$ Hence
$$[e^{i\xi}-1-i\xi]\widehat{f}(\xi)=0$$ which implies $\widehat{f}(\xi)=0$ a.e., hence $f=0.$
Since $I-T$ is injective and not invertible for $1<p\le 2$ then by the Banach inverse mapping theorem the operator $I-T$ cannot be surjective.
Remark
In order to complete the proof it remains to show that $T-I$ is injective for
$p>2.$ The Fourier transform is not available, but perhaps there is a direct way of getting the injectivity. For example equivalently it suffices to show that the range of the adjoint operator $I-T'$ is dense in $L^{p'}.$ Studying the adjoint operator was suggested by @Yaddle
