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I am trying to bound the gain (or operator norm) of a convolutional operator.

For 1-D, applying Young's Inequality is straightforward:

$ ||f * g||_r \le ||f||_p ||g||_q $,

with $ 1 \le p,q,r \le \infty$, $ \frac{1}{p} + \frac{1}{q} = \frac{1}{r} + 1 $ and $ f \in L^p(\mathbb{R}^d), g \in L^p(\mathbb{R}^d) $.

So if I take my conv "filter" (or kernel) to be $f$ and the input to that filter ($\mathbf{x}$) is $g$, using the inequality above I can write:

$ || f * \mathbf{x}||_2 \le ||f||_1 ||\mathbf{x}||_2 $

However, I am unsure about how to generalize this to the 2-D or 3-D case for a filter bank, where the convolutional filters in each dimension are not separable (so one cannot simply apply the 1-D bound iteratively for each dimension).

E.g., for 2-D signals in the continuous setting, the convolution would be written as

$ f(x,y) * g(x,y) = \int \int f(u,v) g(x-u, y-v) du dv $,

where the limits of the integrals are from $-\infty$ to $\infty$.

Is it valid to think of "reshaping" a 2-D or N-D convolution to be in 1-D (something like reshaping a 3x3 matrix to be a 9x1 vector in finite dimensions) and then apply Young's Inequality for finite dimensional vectors?

As a concrete example in finite dimensions: What if there is a filter bank of $K$ convolutional filters, each being a 2-D filter $ w \times h$, that operates on 2-D input tensors of size $ W \times H$ ?

AruniRC
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  • This answer shows Young's Inequality as the consequence of Hölder's Inequality and does not depend on the space being $\mathbb{R}$. – robjohn Oct 02 '17 at 15:33
  • @robjohn so the L1-norm for a convolutional kernel $f$ would simply be $\int \int f^2(x,y) dxdy$ ? And for finite dimensions this is just the squared sum of the elements of the convolution kernel? – AruniRC Oct 02 '17 at 19:54
  • For a convolution operator, the $L^p$ norm is less than or equal to the $L^1$ norm of the kernel. – robjohn Oct 02 '17 at 22:53
  • @robjohn i meant L2-norm in the previous comment. And to confirm, isn't your statement about $L^p$ norm being $\le$ $L^1$ norm for $p \gt 1$ generally true in finite dimensions, not just for convolution kernels? – AruniRC Oct 03 '17 at 14:57
  • What I meant was that $ \left|f\ast g\right|_p \le\left|f\right|_1\left|g\right|_p $ – robjohn Oct 03 '17 at 15:31

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