Convolution norm in $L^1$ and $L^\infty$

Question

Let $f\in L^1(\mathbb{R}^d)$. Show that $T_f:g\mapsto f\star g$ is of norm $\Vert f\Vert_1$ as a linear operator from $L^1(\mathbb{R}^d)$ to $L^1(\mathbb{R}^d)$ and from $L^\infty(\mathbb{R}^d)$ to $L^\infty(\mathbb{R}^d)$.

My work:

Showing that this operator is indeed a continuous linear operator is easy. For $L^1(\mathbb{R}^d)$, we take $(\rho_n)_n$ an approximation of unity, and this yields a sequence in $L^1(\mathbb{R}^d)$ such that $\rho_n\star f\to f$ in $L^1(\mathbb{R}^d)$. Because $\Vert \rho_n\Vert_{L^1}=1$ for all $n$, we are done showing the equality $\Vert T_f\Vert_{\mathcal{L}(L^1(\mathbb{R}^d))}=\Vert f\Vert_{L^1}$.

However, I do not know how to do it for $L^\infty(\mathbb{R}^d)$. We do not always have the above convergence of an approximation of unity in $L^\infty(\mathbb{R}^d)$ (only if $f$ has a representant which is uniformly continuous).

Any hints ? Thanks !

Answer 1

As far as I understand, you are able to prove that the operator $T_t: L^{\infty} \to L^{\infty}$ has norm $\|f\|_{L^1}$ if $f \in L^1$ is uniformly continuous. You can extend the assertion using a density argument.

Let $f \in L^1$ be arbitrary, then there exists a sequence of uniformly continuous functions $(f_n)_{n \in \mathbb{N}} \subseteq L^1$ such that $f_n \to f$ in $L^1$ (e.g. because $C_c(\mathbb{R}^d)$ is dense in $L^1(\mathbb{R}^d)$). Now, by the triangle inequality,

$$\|T_{f_n}(g)\|_{L^{\infty}} = \|(f_n-f)*g+f*g\|_{L^{\infty}} \leq \|f*g\|_{L^{\infty}} + \|(f_n-f)*g\|_{L^{\infty}}.$$

Since

$$\|(f_n-f)*g\|_{L^{\infty}} \leq \|f_n-f\|_{L^1} \|g\|_{L^{\infty}}$$

we find that there exists for any $\epsilon>0$ some $N \in \mathbb{N}$ such that

$$\|T_{f_n}(g)\|_{L^{\infty}} \leq \|f*g\|_{L^{\infty}} + \epsilon\|g\|_{L^{\infty}} = \|T_f(g)\|_{L^{\infty}} + \epsilon \|g\|_{L^{\infty}}\tag{1}$$

for all $n \geq N$. Since $f_n$ is uniformly continuous, there exists for every $k \geq 1$ some $g_k \in L^{\infty}$, $\|g_k\|_{L^{\infty}} \leq 1$, such that $\|T_{f_n}(g_k)\| \geq (\|f_n\|_{L^1}-\frac{1}{k})$. Thus, by $(1)$,

\begin{align*} \sup_{g \in L^{\infty}, \|g\|_{L^{\infty}} \leq 1} \|T_f(g)\| &\geq \sup_{k} \|T_{f_n}(g_k)\|_{L^{\infty}} - \epsilon \\ &\geq \sup_k \left(\|f_n\|_{L^1}-\frac{1}{k}\right)-\epsilon \\ &= \|f_n\|_{L^1}-\epsilon. \end{align*}

Since $n \geq N$ is arbitrary and $\|f_n\|_{L^1} \to \|f\|_{L^1}$, this gives

$$\sup_{g \in L^{\infty}, \|g\|_{L^{\infty}} \leq 1} \|T_f(g)\| \geq \|f\|_{L^1}-\epsilon.$$

Finally, we note that $\epsilon>0$ was arbitrary, and the proof is finished.

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Edit: Here is a way to prove the assertion for $f \in L^1$ continuous.

Let $f \in L^1$ be continuous. If $g \in L^{\infty}$ is continuous, then it follows easily from the dominated convergence theorem that $f*g$ is continuous. For $n \in \mathbb{N}$ let $h_n$ be the piecewise linear function such that $h_n(x)=-1$ for $x \leq - \frac{1}{n}$ and $h_n(x)=1$ for $x \geq \frac{1}{n}$. Set $g_n(x) := h_n(f(-x))$, then $g_n$ (hence $f*g_n$) is continuous and

$$\|f*g_n\|_{L^{\infty}} \geq |f*g_n(0)|. \tag{1}$$

By definition,

$$y h_n(y) \geq 0, \qquad y \in \mathbb{R},$$ and

$$y h_n(y) = |y|, \qquad |y| \geq \frac{1}{n}.$$

This implies

$$(f*g_n)(0) = \int f(x) h_n(f(x)) dx \geq \int_{|f(x)| \geq 1/n} |f(x)| \, dx.$$

Combining this with $(1)$ and letting $n \to \infty$ we get

$$\sup_{\|g\|_{L^{\infty}} \leq 1} \|f*g\|_{L^{\infty}} \geq \|f\|_{L^1}.$$

Answer 2

Here is a different solution based on the following fact:

If $\frac1p+\frac1q=1$, $f\in L_p(\lambda)$ and $g\in L_q(\lambda)$, then $f*g$ is uniformly continuous:

Here is a short proof of this: $1\leq p<\infty$ $$\begin{align} |(f*g)(x+h)-(f*g)(x+k)|&\leq \int|f(x+h-y)-f(x+k-y)||g(y)|\,dy\\ &\leq\|\tau_{-(k-h)}f-f\|_p\|g\|_q \end{align}$$ The conclusion then follows by $L_p$ continuity of the translation operator.

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Back to the OP

Consider $\phi(x)=\frac{\overline{f(x)}}{|f(x)|}$ when $f(x)\neq0$ and $0$ otherwise. The $\phi\in L_\infty$ since $\|\phi\|_\infty=1$. Hence $f*g$ is bounded (by $\|f\|_1$) and uniformly continuous. For $x=0$ $$ |(f*g)(0)|=\int |f| =\|f\|_1$$ Thus $\|f*g\|_\infty=\|f\|_1$